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 ll = long long;
template <int MOD>
struct mint {
 public:
  int val;
  mint(int _val = 0) : val(_val) {
    val = (-MOD <= val && val < MOD) ? val : val % MOD;
    if (val < 0) val += MOD;
  }
  explicit operator int() const { return val; }
  mint operator-() const { return mint(-val); }
  mint& operator+=(mint m) {
    if ((val += m.val) >= MOD) val -= MOD;
    return *this;
  }
  mint& operator-=(mint m) {
    if ((val -= m.val) < 0) val += MOD;
    return *this;
  }
  mint& operator*=(mint m) {
    val = static_cast<int>(static_cast<ll>(val) * m.val % MOD);
    return *this;
  }
  bool operator==(mint m) const { return val == m.val; }
  friend mint pow(mint a, int p) {
    mint r(1);
    for (; p; p >>= 1, a *= a)
      if (p & 1) r *= a;
    return r;
  }
  friend mint operator+(mint a, mint b) { return a += b; }
  friend mint operator-(mint a, mint b) { return a -= b; }
  friend mint operator*(mint a, mint b) { return a *= b; }
};
const int MK = 1e5 + 1;
const int MN = 1e5 + 10;
using mint0 = mint<1000000007>;
const mint0 base0 = mint0(MK);
mint0 pow0[MN];
struct hash {
 public:
  mint0 val0;
  hash(int x = 0) { val0 = mint0(x); }
  bool operator==(hash o) const { return val0 == o.val0; }
  hash& operator+=(hash o) {
    val0 += o.val0;
    return *this;
  }
  hash& operator*=(int o) {
    if (o) val0 *= pow0[o];
    return *this;
  }
  friend hash operator+(hash a, hash b) { return a += b; }
  friend hash operator*(hash a, int b) { return a *= b; }
};
struct node {
 public:
  int s, v;
  char h;
  hash sum;
  int c[2];
};
const int pool_sz = 1 << 23;
int pool_ctr = pool_sz;
node pool[pool_sz];
int nn() {
  assert(pool_ctr > 1);
  return --pool_ctr;
}
void upd(int n) {
  assert(n);
  pool[n].s = pool[pool[n].c[0]].s + pool[pool[n].c[1]].s + 1;
  pool[n].h = std::max(pool[pool[n].c[0]].h, pool[pool[n].c[1]].h) + 1;
  pool[n].sum = hash(pool[n].v) * pool[pool[n].c[0]].s;
  if (pool[n].c[0]) pool[n].sum += pool[pool[n].c[0]].sum;
  if (pool[n].c[1])
    pool[n].sum += pool[pool[n].c[1]].sum * (pool[pool[n].c[0]].s + 1);
}
int nn(int x) {
  int f = nn();
  pool[f].v = x;
  return upd(f), f;
}
int copy(int x) {
  int f = nn();
  pool[f] = pool[x];
  return f;
}
int rot(int n, bool r, bool pst) {
  int o = pst ? copy(pool[n].c[r]) : pool[n].c[r];
  pool[n].c[r] = pool[o].c[!r], pool[o].c[!r] = n;
  return upd(n), upd(o), o;
}
int bal(int n, bool pst) {
  upd(n);
  int b = pool[pool[n].c[0]].h - pool[pool[n].c[1]].h;
  if (b > 1)
    if (pool[pool[pool[n].c[0]].c[0]].h + 1 == pool[pool[n].c[0]].h)
      return rot(n, 0, 0);
    else
      return pool[n].c[0] = rot(pool[n].c[0], 1, pst), rot(n, 0, 0);
  else if (b < -1)
    if (pool[pool[pool[n].c[1]].c[1]].h + 1 == pool[pool[n].c[1]].h)
      return rot(n, 1, 0);
    else
      return pool[n].c[1] = rot(pool[n].c[1], 0, pst), rot(n, 1, 0);
  else
    return n;
}
int merge(int a, int b, bool pst) {
  if (!b) return a;
  if (!a) return b;
  if (pool[a].h < pool[b].h)
    return b = pst ? copy(b) : b, pool[b].c[0] = merge(a, pool[b].c[0], pst),
           bal(b, pst);
  else
    return a = pst ? copy(a) : a, pool[a].c[1] = merge(pool[a].c[1], b, pst),
           bal(a, pst);
}
int mwr(int l, int m, int r, bool pst) {
  int b = pool[l].h - pool[r].h;
  if (b > 1)
    return l = pst ? copy(l) : l, pool[l].c[1] = mwr(pool[l].c[1], m, r, pst),
           bal(l, pst);
  else if (b < -1)
    return r = pst ? copy(r) : r, pool[r].c[0] = mwr(l, m, pool[r].c[0], pst),
           bal(r, pst);
  pool[m].c[0] = l, pool[m].c[1] = r, upd(m);
  return m;
}
struct cut {
 public:
  int l, r;
};
cut split(int n, int k, bool pst) {
  if (!k) return {0, n};
  if (k == pool[n].s) return {n, 0};
  if (pool[pool[n].c[0]].s < k) {
    cut x = split(pool[n].c[1], k - pool[pool[n].c[0]].s - 1, pst);
    return {mwr(pool[n].c[0], pst ? nn(pool[n].v) : n, x.l, pst), x.r};
  } else {
    cut x = split(pool[n].c[0], k, pst);
    return {x.l, mwr(x.r, pst ? nn(pool[n].v) : n, pool[n].c[1], pst)};
  }
}
int splitR(int n, int k) {
  if (k == pool[n].s) return 0;
  if (!k) return n;
  if (pool[pool[n].c[0]].s < k)
    return splitR(pool[n].c[1], k - pool[pool[n].c[0]].s - 1);
  else {
    int x = splitR(pool[n].c[0], k);
    return mwr(x, nn(pool[n].v), pool[n].c[1], 1);
  }
}
hash getval(int n, int k) {
  if (!k) return hash(0);
  if (pool[n].s == k) return pool[n].sum;
  if (pool[pool[n].c[0]].s < k)
    return pool[pool[n].c[0]].sum + hash(pool[n].v) * pool[pool[n].c[0]].s +
           getval(pool[n].c[1], k - pool[pool[n].c[0]].s - 1) *
               (pool[pool[n].c[0]].s + 1);
  else
    return getval(pool[n].c[0], k);
}
void out(int n) {
  if (!n) return;
  out(pool[n].c[0]);
  printf("id: %3d(%d), c:[%3d, %3d], s[%2d], h[%2d], hash: (%d, %d)\n", n,
         pool[n].v, pool[n].c[0], pool[n].c[1], pool[n].s, pool[n].h,
         pool[n].sum.val0, -1);
  out(pool[n].c[1]);
}
int N, K, Q, s[MN];
const int MS = 1 << 18;
struct state {
 public:
  int open, close;
  bool ok;
  state(int open, int close, bool ok) : open(open), close(close), ok(ok) {}
  state(int x = 0) {
    ok = 1;
    close = open = 0;
    if (x < 0) close = nn(-x);
    if (0 < x) open = nn(x);
  }
  void fail() { open = close = 0, ok = 0; }
  state& operator+=(const state& o) {
    if (ok) {
      if (!o.ok) return fail(), *this;
      int b = pool[open].s - pool[o.close].s;
      if (b > 0) {
        cut x;
        if (!o.close)
          open = merge(o.open, open, 1);
        else if (pool[(x = split(open, pool[o.close].s, 1)).l].sum ==
                 pool[o.close].sum)
          open = merge(o.open, x.r, 1);
        else
          return fail(), *this;
      } else if (b < 0) {
        cut x;
        if (!open)
          open = o.open, close = merge(close, o.close, 1);
        else if (pool[(x = split(o.close, pool[open].s, 1)).l].sum ==
                 pool[open].sum)
          close = merge(close, x.r, 1), open = o.open;
        else
          return fail(), *this;
      } else if (!open || pool[open].sum == pool[o.close].sum)
        open = o.open;
      else
        return fail(), *this;
    }
    return *this;
  }
  friend state operator+(state a, state b) { return a += b; }
};
state st[MS];
int cur;
void build(int n, int l, int r) {
  if (r - l > 1) {
    int m = l + (r - l) / 2;
    build(n << 1, l, m);
    build(n << 1 | 1, m, r);
    st[n] = st[n << 1] + st[n << 1 | 1];
  } else
    st[n] = state(s[l]);
}
void upd(int n, int l, int r, int x) {
  if (r - l > 1) {
    int m = l + (r - l) / 2;
    if (x < m)
      upd(n << 1, l, m, x);
    else
      upd(n << 1 | 1, m, r, x);
    st[n] = st[n << 1] + st[n << 1 | 1];
  } else
    st[n] = state(s[l]);
}
void query(int n, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    if (~cur) {
      if (!st[n].ok)
        cur = -1;
      else if (pool[st[n].close].s > pool[cur].s)
        cur = -1;
      else if (!cur)
        cur = st[n].open;
      else if (getval(cur, pool[st[n].close].s) == pool[st[n].close].sum)
        cur = merge(st[n].open, splitR(cur, pool[st[n].close].s), 1);
      else
        cur = -1;
    }
  } else {
    int m = l + (r - l) / 2;
    if (ql < m) query(n << 1, l, m, ql, qr);
    if (m < qr) query(n << 1 | 1, m, r, ql, qr);
  }
}
int main(void) {
  pow0[0] = mint0(1);
  for (int i = 0; i + 1 < MN; ++i) pow0[i + 1] = pow0[i] * base0;
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; ++i) scanf("%d", s + i);
  build(1, 0, N);
  scanf("%d", &Q);
  for (int i = 0, x, y, z; i < Q; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1)
      s[--y] = z, upd(1, 0, N, y);
    else {
      cur = 0;
      query(1, 0, N, --y, z);
      if (cur)
        printf("No\n");
      else
        printf("Yes\n");
    }
    if (pool_ctr < 1000) {
      memset(pool, 0, sizeof(pool));
      pool_ctr = pool_sz;
      build(1, 0, N);
    }
  }
  return 0;
}