Problem: This is the hard version of the problem. You can find the easy version in the Div. 2 contest. Both versions only differ in the number of times you can ask your friend to taste coffee.

This is an interactive problem.

You're considering moving to another city, where one of your friends already lives. There are n cafés in this city, where n is a power of two. The i-th café produces a single variety of coffee a_i. 

As you're a coffee-lover, before deciding to move or not, you want to know the number d of distinct varieties of coffees produced in this city.

You don't know the values a_1, …, a_n. Fortunately, your friend has a memory of size k, where k is a power of two.

Once per day, you can ask him to taste a cup of coffee produced by the café c, and he will tell you if he tasted a similar coffee during the last k days.

You can also ask him to take a medication that will reset his memory. He will forget all previous cups of coffee tasted. You can reset his memory at most 30\ 000 times.

More formally, the memory of your friend is a queue S. Doing a query on café c will: 

  * Tell you if a_c is in S; 
  * Add a_c at the back of S; 
  * If |S| > k, pop the front element of S. 



Doing a reset request will pop all elements out of S.

Your friend can taste at most (3n^2)/(2k) cups of coffee in total. Find the diversity d (number of distinct values in the array a).

Note that asking your friend to reset his memory does not count towards the number of times you ask your friend to taste a cup of coffee.

In some test cases the behavior of the interactor is adaptive. It means that the array a may be not fixed before the start of the interaction and may depend on your queries. It is guaranteed that at any moment of the interaction, there is at least one array a consistent with all the answers given so far.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 1024, k and n are powers of two).

It is guaranteed that (3n^2)/(2k) ≤ 15\ 000.

Interaction

You begin the interaction by reading n and k.

  * To ask your friend to taste a cup of coffee produced by the café c, in a separate line output

? c

Where c must satisfy 1 ≤ c ≤ n. Don't forget to flush, to get the answer.

In response, you will receive a single letter Y (yes) or N (no), telling you if variety a_c is one of the last k varieties of coffee in his memory.

  * To reset the memory of your friend, in a separate line output the single letter R in upper case. You can do this operation at most 30\ 000 times.
  * When you determine the number d of different coffee varieties, output

! d




In case your query is invalid, you asked more than (3n^2)/(2k) queries of type ? or you asked more than 30\ 000 queries of type R, the program will print the letter E and will finish interaction. You will receive a Wrong Answer verdict. Make sure to exit immediately to avoid getting other verdicts.

After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++; 
  * System.out.flush() in Java; 
  * flush(output) in Pascal; 
  * stdout.flush() in Python; 
  * see documentation for other languages. 



Hack format

The first line should contain the word fixed

The second line should contain two integers n and k, separated by space (1 ≤ k ≤ n ≤ 1024, k and n are powers of two).

It must hold that (3n^2)/(2k) ≤ 15\ 000.

The third line should contain n integers a_1, a_2, …, a_n, separated by spaces (1 ≤ a_i ≤ n).

Examples

Input


4 2
N
N
Y
N
N
N
N


Output


? 1
? 2
? 3
? 4
R
? 4
? 1
? 2
! 3


Input


8 8
N
N
N
N
Y
Y


Output


? 2
? 6
? 4
? 5
? 2
? 5
! 6

Note

In the first example, the array is a = [1, 4, 1, 3]. The city produces 3 different varieties of coffee (1, 3 and 4).

The successive varieties of coffee tasted by your friend are 1, 4, 1, 3, 3, 1, 4 (bold answers correspond to Y answers). Note that between the two ? 4 asks, there is a reset memory request R, so the answer to the second ? 4 ask is N. Had there been no reset memory request, the answer to the second ? 4 ask is Y.

In the second example, the array is a = [1, 2, 3, 4, 5, 6, 6, 6]. The city produces 6 different varieties of coffee.

The successive varieties of coffee tasted by your friend are 2, 6, 4, 5, 2, 5.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 60;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  long long to, cap, rev;
  Edge(long long _to, long long _cap, long long _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, long long from, long long to,
              long long cap, bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size()));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (int(i) = (int)(0); i < (int)(used.size()); i++) used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  for (int(i) = (int)(0); i < (int)(d.size()); i++) negative[i] = false;
  d[s] = 0;
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (int(k) = (int)(0); k < (int)(G.size() - 1); k++) {
    for (int(i) = (int)(0); i < (int)(G.size()); i++) {
      for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (int(i) = (int)(0); i < (int)(G[a.second].size()); i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) d[i].resize(G.size());
  for (int(i) = (int)(0); i < (int)(d.size()); i++) {
    for (int(j) = (int)(0); j < (int)(d[i].size()); j++) {
      d[i][j] = INF;
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G[i].size()); j++) {
      d[i][G[i][j].to] = G[i][j].cap;
    }
  }
  for (int(i) = (int)(0); i < (int)(G.size()); i++) {
    for (int(j) = (int)(0); j < (int)(G.size()); j++) {
      for (int(k) = (int)(0); k < (int)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<int>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (int(i) = (int)(0); i < (int)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  lca() {}
  lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
void visit(const vector<vector<Edge>>& g, int v, vector<vector<long long>>& scc,
           stack<long long>& S, vector<long long>& inS, vector<long long>& low,
           vector<long long>& num, int& time) {
  low[v] = num[v] = ++time;
  S.push(v);
  inS[v] = true;
  for (decltype((g[v]).begin()) e = (g[v]).begin(); e != (g[v]).end(); ++e) {
    int w = e->to;
    if (num[w] == 0) {
      visit(g, w, scc, S, inS, low, num, time);
      low[v] = min(low[v], low[w]);
    } else if (inS[w])
      low[v] = min(low[v], num[w]);
  }
  if (low[v] == num[v]) {
    scc.push_back(vector<long long>());
    while (1) {
      int w = S.top();
      S.pop();
      inS[w] = false;
      scc.back().push_back(w);
      if (v == w) break;
    }
  }
}
void stronglyConnectedComponents(const vector<vector<Edge>>& g,
                                 vector<vector<long long>>& scc) {
  const int n = g.size();
  vector<long long> num(n), low(n);
  stack<long long> S;
  vector<long long> inS(n);
  int time = 0;
  for (int(u) = (int)(0); u < (int)(n); u++)
    if (num[u] == 0) visit(g, u, scc, S, inS, low, num, time);
}
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
  int numSet() { return num; }
};
class SumSegTree {
 private:
  long long _sum(long long a, long long b, long long k, long long l,
                 long long r) {
    if (r <= a || b <= l) return 0;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r);
      return s1 + s2;
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  SumSegTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, 0);
  }
  void add(long long i, long long x) {
    i += n - 1;
    dat[i] += x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] += x;
    }
  }
  long long sum(long long a, long long b) { return _sum(a, b, 0, 0, n); }
};
class RmqTree {
 private:
  long long _find(long long a, long long b, long long k, long long l,
                  long long r) {
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long s1 = _find(a, b, 2 * k + 1, l, (l + r) / 2);
      long long s2 = _find(a, b, 2 * k + 2, (l + r) / 2, r);
      return min(s1, s2);
    }
  }

 public:
  long long n, height;
  vector<long long> dat;
  RmqTree(long long _n) {
    n = 1;
    height = 1;
    while (n < _n) {
      n *= 2;
      height++;
    }
    dat = vector<long long>(2 * n - 1, INF);
  }
  void update(long long i, long long x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  long long find(long long a, long long b) { return _find(a, b, 0, 0, n); }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
vector<long long> lis_fast(const vector<long long>& a) {
  const long long n = a.size();
  vector<long long> A(n, INT_MAX);
  vector<long long> id(n);
  for (int i = 0; i < n; ++i) {
    id[i] = distance(A.begin(), lower_bound(A.begin(), A.end(), a[i]));
    A[id[i]] = a[i];
  }
  long long m = *max_element(id.begin(), id.end());
  vector<long long> b(m + 1);
  for (int i = n - 1; i >= 0; --i)
    if (id[i] == m) b[m--] = a[i];
  return b;
}
bool z_algorithm(string& str, vector<int>& z, long long s) {
  const int L = str.size();
  z.resize(str.size());
  for (int i = 1, left = 0, right = 0; i < L; i++) {
    if (i > right) {
      left = right = i;
      for (; right < L && str[right - left] == str[right]; right++)
        ;
      z[i] = right - left;
      right--;
    } else {
      int k = i - left;
      if (z[k] < right - i + 1) {
        z[i] = z[k];
      } else {
        left = i;
        for (; right < L && str[right - left] == str[right]; right++)
          ;
        z[i] = right - left;
        right--;
      }
    }
    if (z[i] == s) return true;
  }
  return false;
}
bool z_algorithm(string& str, vector<int>& z) {
  const int L = str.size();
  z.resize(str.size());
  for (int i = 1, left = 0, right = 0; i < L; i++) {
    if (i > right) {
      left = right = i;
      for (; right < L && str[right - left] == str[right]; right++)
        ;
      z[i] = right - left;
      right--;
    } else {
      int k = i - left;
      if (z[k] < right - i + 1) {
        z[i] = z[k];
      } else {
        left = i;
        for (; right < L && str[right - left] == str[right]; right++)
          ;
        z[i] = right - left;
        right--;
      }
    }
  }
  return true;
}
struct RollingHash {
  static const int base1 = 1007, base2 = 2009;
  static const int mod1 = 1000000007, mod2 = 1000000009;
  vector<long long> hash1, hash2, power1, power2;
  RollingHash(const string& S) {
    int n = (int)S.size();
    hash1.assign(n + 1, 0);
    hash2.assign(n + 1, 0);
    power1.assign(n + 1, 1);
    power2.assign(n + 1, 1);
    for (int i = 0; i < n; ++i) {
      hash1[i + 1] = (hash1[i] * base1 + S[i]) % mod1;
      hash2[i + 1] = (hash2[i] * base2 + S[i]) % mod2;
      power1[i + 1] = (power1[i] * base1) % mod1;
      power2[i + 1] = (power2[i] * base2) % mod2;
    }
  }
  inline pair<long long, long long> get(int l, int r) const {
    long long res1 = hash1[r] - hash1[l] * power1[r - l] % mod1;
    if (res1 < 0) res1 += mod1;
    long long res2 = hash2[r] - hash2[l] * power2[r - l] % mod2;
    if (res2 < 0) res2 += mod2;
    return {res1, res2};
  }
  inline int getLCP(int a, int b) const {
    int len = min((int)hash1.size() - a, (int)hash1.size() - b);
    int low = 0, high = len;
    while (high - low > 1) {
      int mid = (low + high) >> 1;
      if (get(a, a + mid) != get(b, b + mid))
        high = mid;
      else
        low = mid;
    }
    return low;
  }
};
long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1LL;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> inv;
  long long mod;
  long long mod_inv(long long x) {
    long long n = mod - 2LL;
    long long res = 1LL;
    while (n > 0) {
      if (n & 1) res = res * x % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return res;
  }
  long long nCr(long long n, long long r) {
    return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long n, long long _mod) {
    mod = _mod;
    fact.resize(n + 1);
    fact[0] = 1;
    for (int(i) = (int)(0); i < (int)(n); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    inv.resize(n + 1);
    inv[n] = mod_inv(fact[n]);
    for (int i = n; i > 0; i--) {
      inv[i - 1] = inv[i] * i % mod;
    }
  }
};
long long gcd(long long m, long long n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long lcm(long long m, long long n) { return m / gcd(m, n) * n; }
vector<vector<long long>> mIdentity(long long n) {
  vector<vector<long long>> A(n, vector<long long>(n));
  for (int i = 0; i < n; ++i) A[i][i] = 1;
  return A;
}
vector<vector<long long>> mMul(const vector<vector<long long>>& A,
                               const vector<vector<long long>>& B) {
  vector<vector<long long>> C(A.size(), vector<long long>(B[0].size()));
  for (int i = 0; i < C.size(); ++i)
    for (int j = 0; j < C[i].size(); ++j)
      for (int k = 0; k < A[i].size(); ++k)
        (C[i][j] += (A[i][k] % 1000000007) * (B[k][j] % 1000000007)) %=
            1000000007;
  return C;
}
vector<vector<long long>> mPow(const vector<vector<long long>>& A,
                               long long e) {
  return e == 0       ? mIdentity(A.size())
         : e % 2 == 0 ? mPow(mMul(A, A), e / 2)
                      : mMul(A, mPow(A, e - 1));
}
template <class T>
class RectangleSum {
 public:
  vector<vector<T>> sum;
  T GetSum(int left, int right, int top, int bottom) {
    T res = sum[bottom][right];
    if (left > 0) res -= sum[bottom][left - 1];
    if (top > 0) res -= sum[top - 1][right];
    if (left > 0 && top > 0) res += sum[top - 1][left - 1];
    return res;
  }
  RectangleSum(const vector<vector<T>>& s, int h, int w) {
    sum.resize(h);
    for (int i = 0; i < h; i++) sum[i].resize(w, 0);
    for (int y = 0; y < h; y++) {
      for (int x = 0; x < w; x++) {
        sum[y][x] = s[y][x];
        if (y > 0) sum[y][x] += sum[y - 1][x];
        if (x > 0) sum[y][x] += sum[y][x - 1];
        if (y > 0 && x > 0) sum[y][x] -= sum[y - 1][x - 1];
      }
    }
  }
};
long long _garner(vector<long long>& xs, vector<long long>& mods) {
  int M = xs.size();
  vector<long long> coeffs(M, 1), constants(M, 0);
  for (int i = 0; i < M - 1; ++i) {
    long long mod_i = mods[i];
    long long v = (xs[i] - constants[i] + mod_i) % mod_i;
    v = (v * mod_pow(coeffs[i], mod_i - 2, mod_i)) % mod_i;
    for (int j = i + 1; j < M; j++) {
      long long mod_j = mods[j];
      constants[j] = (constants[j] + coeffs[j] * v) % mod_j;
      coeffs[j] = (coeffs[j] * mod_i) % mod_j;
    }
  }
  return constants.back();
}
template <typename T>
inline void bit_reverse(vector<T>& a) {
  int n = a.size();
  int i = 0;
  for (int j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(a[i], a[j]);
  }
}
template <long long mod, long long primitive_root>
class NTT {
 public:
  long long get_mod() { return mod; }
  void _ntt(vector<long long>& a, int sign) {
    const int n = a.size();
    assert((n ^ (n & -n)) == 0);
    const long long g = primitive_root;
    long long tmp = (mod - 1) * mod_pow(n, mod - 2, mod) % mod;
    long long h = mod_pow(g, tmp, mod);
    if (sign == -1) h = mod_pow(h, mod - 2, mod);
    bit_reverse(a);
    for (int m = 1; m < n; m <<= 1) {
      const int m2 = 2 * m;
      long long _base = mod_pow(h, n / m2, mod);
      long long _w = 1;
      for (int x = 0; x < m; ++x) {
        for (int s = x; s < n; s += m2) {
          long long u = a[s];
          long long d = (a[s + m] * _w) % mod;
          a[s] = (u + d) % mod;
          a[s + m] = (u - d + mod) % mod;
        }
        _w = (_w * _base) % mod;
      }
    }
  }
  void ntt(vector<long long>& input) { _ntt(input, 1); }
  void intt(vector<long long>& input) {
    _ntt(input, -1);
    const long long n_inv = mod_pow(input.size(), mod - 2, mod);
    for (auto& x : input) x = (x * n_inv) % mod;
  }
  vector<long long> convolution(const vector<long long>& a,
                                const vector<long long>& b) {
    int result_size = a.size() + b.size() - 1;
    int n = 1;
    while (n < result_size) n <<= 1;
    vector<long long> _a = a, _b = b;
    _a.resize(n, 0);
    _b.resize(n, 0);
    ntt(_a);
    ntt(_b);
    for (int i = 0; i < n; ++i) _a[i] = (_a[i] * _b[i]) % mod;
    intt(_a);
    _a.resize(result_size);
    return _a;
  }
};
vector<long long> convolution_ntt(vector<long long>& a, vector<long long>& b,
                                  long long mod = 1224736769LL) {
  for (auto& x : a) x %= mod;
  for (auto& x : b) x %= mod;
  long long maxval = max(a.size(), b.size()) *
                     *max_element(a.begin(), a.end()) *
                     *max_element(b.begin(), b.end());
  if (maxval < 1224736769) {
    NTT<1224736769, 3> ntt3;
    return ntt3.convolution(a, b);
  }
  NTT<167772161, 3> ntt1;
  NTT<469762049, 3> ntt2;
  NTT<1224736769, 3> ntt3;
  vector<long long> x1 = ntt1.convolution(a, b);
  vector<long long> x2 = ntt2.convolution(a, b);
  vector<long long> x3 = ntt3.convolution(a, b);
  vector<long long> ret(x1.size());
  vector<long long> mods{167772161, 469762049, 1224736769, mod};
  for (int i = 0; i < x1.size(); ++i) {
    vector<long long> xs{x1[i], x2[i], x3[i], 0};
    ret[i] = _garner(xs, mods);
  }
  return ret;
}
int popcount3(int x) {
  x = (x & 0x55555555) + (x >> 1 & 0x55555555);
  x = (x & 0x33333333) + (x >> 2 & 0x33333333);
  x = (x & 0x0F0F0F0F) + (x >> 4 & 0x0F0F0F0F);
  x = (x & 0x00FF00FF) + (x >> 8 & 0x00FF00FF);
  x = (x & 0x0000FFFF) + (x >> 16 & 0x0000FFFF);
  return x;
}
template <typename T>
void rowReduction(vector<T> mat, vector<T>& basis) {
  for (auto e : mat) {
    for (auto b : basis) chmin(e, e ^ b);
    if (e) basis.push_back(e);
  }
  sort((basis).begin(), (basis).end(), greater<T>());
}
struct SegT {
 private:
  int n;
  vector<long long> node, lazy;
  const long long init_c = 0;

 public:
  SegT(int sz) {
    n = 1;
    while (n < sz) n <<= 1;
    node.resize(2 * n - 1, 0);
    lazy.resize(2 * n - 1, 0);
  }
  long long f(long long a, long long b) { return max(a, b); }
  void eval(int k, int l, int r) {
    node[k] += lazy[k];
    if (r - l > 1) {
      lazy[2 * k + 1] += lazy[k];
      lazy[2 * k + 2] += lazy[k];
    }
    lazy[k] = 0;
  }
  void add(long long x, int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (r <= a || b <= l) return;
    if (a <= l && r <= b) {
      lazy[k] += x;
      eval(k, l, r);
    } else {
      add(x, a, b, k * 2 + 1, l, (l + r) / 2);
      add(x, a, b, k * 2 + 2, (l + r) / 2, r);
      node[k] = f(node[k * 2 + 1], node[k * 2 + 2]);
    }
  }
  long long query(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (r <= a || b <= l) return init_c;
    if (a <= l && r <= b)
      return node[k];
    else {
      long long vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      long long vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return f(vl, vr);
    }
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> rest(n, 1);
  if (k != 1) k /= 2;
  for (int(i) = (int)(0); i < (int)(n / k); i++) {
    for (int(j) = (int)(0); j < (int)(k); j++) {
      cout << "? " << k * i + j + 1 << endl;
      char in;
      cin >> in;
      if (in == 'Y') rest[k * i + j] = 0;
    }
    cout << "R" << endl;
  }
  for (int(i) = (int)(0); i < (int)(n / k); i++) {
    for (int(j) = (int)(i + 1); j < (int)(n / k); j++) {
      for (int(a) = (int)(0); a < (int)(k); a++) {
        cout << "? " << k * i + a + 1 << endl;
        char in;
        cin >> in;
      }
      for (int(a) = (int)(0); a < (int)(k); a++) {
        cout << "? " << k * j + a + 1 << endl;
        char in;
        cin >> in;
        if (in == 'Y') rest[k * j + a] = 0;
      }
      cout << "R" << endl;
    }
  }
  long long ans = 0;
  for (int(i) = (int)(0); i < (int)(n); i++)
    if (rest[i] == 1) ans++;
  cout << "! " << ans << endl;
  return 0;
}