Problem: Nastya is a competitive programmer, but she is only studying now. Recently, Denis told her about the way to check if the string is correct bracket sequence. After that, unexpectedly, Nastya came up with a much more complex problem that Denis couldn't solve. Can you solve it? 

A string s is given. It consists of k kinds of pairs of brackets. Each bracket has the form t — it is an integer, such that 1 ≤ |t| ≤ k. If the bracket has a form t, then: 

  * If t > 0, then it's an opening bracket of the type t. 
  * If t < 0, then it's a closing bracket of the type -t. 



Thus, there are k types of pairs of brackets in total. 

The queries need to be answered: 

  1. Replace the bracket at position i with the bracket of the form t. 
  2. Check if the substring from the l-th to r-th position (including) is the correct bracket sequence. 

Recall the definition of the correct bracket sequence: 
  * An empty sequence is correct. 
  * If A and B are two correct bracket sequences, then their concatenation "A B" is also correct bracket sequence. 
  * If A is the correct bracket sequence, c (1 ≤ c ≤ k) is a type of brackets, then the sequence "c A -c" is also correct bracket sequence. 

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — length of string and k (1 ≤ k ≤ n) — the number of kinds of pairs of brackets.

The second line contains string s of length n — n integers s_1, s_2, …, s_n (1 ≤ |s_i| ≤ k)

The third line contains single integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the following q lines describes the queries: 

  * 1 i t - query of the 1 type (1 ≤ i ≤ n, 1 ≤ |t| ≤ k). 
  * 2 l r - query of the 2 type (1 ≤ l ≤ r ≤ n). 

Output

For each query of 2 type, output "Yes" if the substring from the query is the correct bracket sequence and "No", otherwise.

All letters can be displayed in any case.

Examples

Input


2 1
1 -1
1
2 1 2


Output


Yes


Input


2 2
1 -2
1
2 1 2


Output


No


Input


6 2
1 2 -2 -1 1 -1
3
2 1 6
2 1 4
2 2 5


Output


Yes
Yes
No


Input


2 2
-1 1
4
2 1 2
1 1 1
1 2 -1
2 1 2


Output


No
Yes

Note

In the fourth test, initially, the string is not a correct bracket sequence, so the answer to the first query is "No". After two changes it will be equal to "1 -1", so it is a correct bracket sequence and the answer to the fourth query is "Yes".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  static char _c;
  static bool _f;
  x = 0;
  _f = 0;
  _c = getchar();
  while (_c < '0' || '9' < _c) {
    if (_c == '-') _f = true;
    _c = getchar();
  }
  while ('0' <= _c && _c <= '9') {
    x = (x << 1) + (x << 3) + (_c & 15);
    _c = getchar();
  }
  if (_f) x = -x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void Min(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
inline void Max(T &x, T y) {
  if (x < y) x = y;
}
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double pi = (double)acos(-1.0);
const double eps = (double)1e-8;
const int e5 = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;
template <typename T0, typename T1>
inline void depair(T0 &x, T1 &y, pair<T0, T1> &p) {
  x = p.first, y = p.second;
}
inline int sig(double x) { return x < -eps ? -1 : eps < x; }
long long fp(long long a, long long n, long long mod = MOD) {
  if (n < 0) a = fp(a, mod - 2, mod), n = -n;
  long long res = 1;
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) res = res * a % mod;
  return res;
}
struct Mint {
  int x;
  Mint() { x = 0; }
  Mint(int _x) : x(_x) {
    if (x < 0 || x >= MOD) x = (x % MOD + MOD) % MOD;
  }
  Mint(long long _x) : x(_x) {
    if (x < 0 || x >= MOD) x = (x % MOD + MOD) % MOD;
  }
  Mint operator-() const { return Mint(MOD - x); }
  Mint operator+(const Mint &rhs) const {
    return Mint(x + rhs.x >= MOD ? x + rhs.x - MOD : x + rhs.x);
  }
  Mint operator-(const Mint &rhs) const {
    return Mint(x - rhs.x < 0 ? x - rhs.x + MOD : x - rhs.x);
  }
  Mint operator*(const Mint &rhs) const {
    return Mint((long long)x * rhs.x % MOD);
  }
  Mint operator/(const Mint &rhs) const {
    return Mint(x * fp(rhs.x, -1) % MOD);
  }
  Mint &operator+=(const Mint &rhs) {
    x += rhs.x;
    if (x >= MOD) x -= MOD;
    return *this;
  }
  Mint &operator*=(const Mint &rhs) {
    x = ((long long)x * rhs.x) % MOD;
    return *this;
  }
  bool operator==(const Mint &rhs) const { return x == rhs.x; }
  bool operator!=(const Mint &rhs) const { return x != rhs.x; }
  friend ostream &operator<<(ostream &out, const Mint &rhs) {
    return out << rhs.x;
  }
  friend istream &operator>>(istream &in, Mint &rhs) { return in >> rhs.x; }
};
const int maxn = (int)1e5 + 20;
const int maxm = (int)1e6 + 20;
const int BASE = 998244353;
int n, k;
int a[maxn];
long long po[maxn];
long long inv[maxn];
struct Block {
  int fail;
  vector<int> arr;
  vector<int> Lpart, Rpart;
  vector<long long> Lhs, Rhs;
  int Ridx;
  void init(vector<int> _arr) { arr = _arr; }
  void build() {
    vector<int> s;
    for (auto x : arr) {
      if (s.size() != 0 && s.back() == -x && x < 0)
        s.pop_back();
      else
        s.push_back(x);
    }
    fail = 0;
    for (int i = 0; i < (int)s.size() - 1; i++) {
      if (s[i] > 0 && s[i + 1] < 0) {
        fail = 1;
        break;
      }
    }
    if (fail) return;
    int i;
    for (i = 0; i < (int)s.size(); i++)
      if (s[i] > 0) break;
    Lpart = vector<int>(s.begin(), s.begin() + i);
    Rpart = vector<int>(s.begin() + i, s.end());
    reverse(Rpart.begin(), Rpart.end());
    for (auto &x : Lpart) x = -x;
    Lhs = vector<long long>(Lpart.size());
    Rhs = vector<long long>(Rpart.size());
    for (int i = 0; i < Lpart.size(); i++) {
      if (i)
        Lhs[i] = (Lhs[i - 1] + Lpart[i] * po[i]) % MOD;
      else
        Lhs[i] = Lpart[i];
    }
    for (int i = 0; i < Rpart.size(); i++) {
      if (i)
        Rhs[i] = (Rhs[i - 1] + Rpart[i] * po[i]) % MOD;
      else
        Rhs[i] = Rpart[i];
    }
  }
  void change(int pos, int val) {
    arr[pos] = val;
    build();
  }
};
void work() {
  read(n, k);
  po[0] = 1;
  for (int i = 1; i <= n; i++) po[i] = po[i - 1] * BASE % MOD;
  inv[n] = fp(po[n], -1);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * BASE % MOD;
  for (int i = 0; i < n; i++) read(a[i]);
  int B = min(n, 2 * (int)sqrt(n) + 1);
  int nn = (n + B - 1) / B;
  vector<Block> bk(nn);
  for (int i = 0; i < nn; i++)
    bk[i].init(vector<int>(a + i * B, a + min((i + 1) * B, n)));
  for (int i = 0; i < nn; i++) bk[i].build();
  int Q;
  read(Q);
  while (Q--) {
    int op, l, r;
    read(op, l, r);
    l--;
    r--;
    int lb = l / B, lp = l % B, rb = r / B, rp = r % B;
    if (op == 1) {
      bk[lb].change(lp, r + 1);
      a[l] = r + 1;
      continue;
    }
    if (lb == rb) {
      Block tmp;
      tmp.init(vector<int>(a + l, a + r + 1));
      tmp.build();
      if (tmp.fail || (tmp.Lhs.size() != 0 && tmp.Lhs.back() != 0) ||
          (tmp.Rhs.size() != 0 && tmp.Rhs.back() != 0))
        cout << "No" << endl;
      else
        cout << "Yes" << endl;
    } else {
      Block L, R;
      L.init(vector<int>(a + l, a + (lb + 1) * B));
      R.init(vector<int>(a + rb * B, a + r + 1));
      L.build();
      R.build();
      if ((L.Lhs.size() != 0 && L.Lhs.back() != 0) ||
          (R.Rhs.size() != 0 && R.Rhs.back() != 0) || L.fail || R.fail) {
        cout << "No" << endl;
        continue;
      }
      int fail = 0;
      for (int i = lb + 1; i < rb; i++) {
        fail |= bk[i].fail;
        bk[i].Ridx = 0;
      }
      L.Ridx = 0;
      stack<Block *> stk;
      if (L.Rpart.size()) stk.push(&L);
      function<long long(vector<long long> &, int, int)> getHash =
          [&](vector<long long> &Hs, int L, int R) -> long long {
        long long res = Hs[R];
        if (L > 0) res = (res - Hs[L - 1] + MOD) * inv[L] % MOD;
        return res;
      };
      function<int(Block *)> add = [&](Block *t) -> int {
        int cur = 0;
        while (stk.size() && cur < t->Lpart.size()) {
          Block *o = stk.top();
          int sz = min(t->Lpart.size() - cur, o->Rpart.size() - o->Ridx);
          if (getHash(o->Rhs, o->Ridx, o->Ridx + sz - 1) ==
              getHash(t->Lhs, cur, cur + sz - 1)) {
            o->Ridx += sz;
            cur += sz;
          } else {
            return -1;
          }
          if (o->Ridx == o->Rhs.size()) stk.pop();
        }
        if (cur != t->Lpart.size()) return -1;
        return 0;
      };
      for (int i = lb + 1; i < rb; i++) {
        if (add(&bk[i]) == -1) {
          fail = 1;
          break;
        }
        if (bk[i].Rpart.size()) stk.push(&bk[i]);
      }
      if (fail) {
        cout << "No" << endl;
        continue;
      }
      if (add(&R) == -1) {
        cout << "No" << endl;
      } else {
        if (stk.size())
          cout << "No" << endl;
        else
          cout << "Yes" << endl;
      }
    }
  }
}
int main(int argc, char **argv) {
  int tc = 1;
  for (int ca = 1; ca <= tc; ca++) {
    work();
  }
  return 0;
}