Problem: There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.

You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).

Your goal is to walk exactly s units of distance in total.

If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.

Input

The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.

Output

If you cannot perform k moves with total walking distance equal to s, print "NO".

Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.

For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.

Examples

Input

10 2 15


Output

YES
10 4 


Input

10 9 45


Output

YES
10 1 10 1 2 1 2 1 6 


Input

10 9 81


Output

YES
10 1 10 1 10 1 10 1 10 


Input

10 9 82


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<ll>;
using VV = vector<VI>;
using VS = vector<string>;
using PII = pair<ll, ll>;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll SUM(VI& V) { return accumulate((V).begin(), (V).end(), 0LL); }
ll MIN(VI& V) { return *min_element((V).begin(), (V).end()); }
ll MAX(VI& V) { return *max_element((V).begin(), (V).end()); }
void print_vector(VI& V) {
  ll n = V.size();
  for (ll i = (0); i < (n); ++i) {
    if (i) cout << ' ';
    cout << V[i];
  }
  cout << endl;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
  ll g = gcd(a, b);
  return a / g * b;
}
using ld = long double;
using inverse_priority_queue = priority_queue<ll, vector<ll>, greater<ll> >;
int popcount(ll t) { return __builtin_popcountll(t); }
void no() {
  cout << ("No") << '\n';
  exit(0);
}
void yes() {
  cout << ("Yes") << '\n';
  exit(0);
}
const ll mod = 1e9 + 7;
const ll inf = 1e18;
const double PI = acos(-1);
ll ceil_div(ll a, ll b) { return (a + b - 1) / b; }
ll string_to_ll(string s) { return atoll(s.c_str()); }
struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};
void solve() {
  ll N;
  cin >> N;
  VI A(N);
  for (ll i = (0); i < (N); ++i) {
    cin >> A[i];
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N, K, S;
  cin >> N >> K >> S;
  if (S < K) {
    cout << ("NO") << '\n';
    return 0;
  }
  ll len = N - 1;
  ll ma = len * K;
  if (ma < S) {
    cout << ("NO") << '\n';
    return 0;
  } else {
    cout << ("YES") << '\n';
  }
  VI D;
  while (S) {
    if (S >= len) {
      D.push_back(len);
      S -= len;
    } else {
      D.push_back(S);
      S = 0;
    }
  }
  42;
  ll one = 0;
  while (((int)(D).size()) < K) {
    ll diff = K - ((int)(D).size());
    ll v = D.back();
    D.pop_back();
    if (v - 1 >= diff) {
      one += diff;
      v -= diff;
      D.push_back(v);
      K = 0;
    } else {
      one += v;
      K -= v;
    }
  }
  while (one--) D.push_back(1);
  42;
  bool right = true;
  ll now = 1;
  VI Ans;
  for (ll d : D) {
    if (right) {
      now += d;
    } else {
      now -= d;
    }
    Ans.push_back(now);
    right = 1 - right;
  }
  print_vector(Ans);
  return 0;
}