Let a_1, …, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).

The weight of a is defined as the maximum number of elements you can remove.

You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the length of the array and the number of queries.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — elements of the array.

The i-th of the next q lines contains two integers x and y (x, y ≥ 0 and x+y < n).

Output

Print q lines, i-th line should contain a single integer — the answer to the i-th query.

Examples

Input


13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12


Output


5
11
6
1
0


Input


5 2
1 4 1 2 4
0 0
1 0


Output


2
0

Note

Explanation of the first query:

After making first x = 3 and last y = 1 elements impossible to remove, a becomes [×, ×, ×, 9, 5, 4, 6, 5, 7, 8, 3, 11, ×] (we represent 14 as × for clarity).

Here is a strategy that removes 5 elements (the element removed is colored in red):

  * [×, ×, ×, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, ×] 
  * [×, ×, ×, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, ×] 
  * [×, ×, ×, 9, 4, \color{red}{6}, 5, 7, 8, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 7, \color{red}{8}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, \color{red}{7}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 3, ×] (final state) 



It is impossible to remove more than 5 elements, hence the weight is 5.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 60;
struct Tree {
  int val[maxn];
  int N;
  inline int lowbit(int x) { return x & (-x); }
  void init(int n) {
    N = n;
    for (int i = 1; i <= n; i++) {
      val[i] = 0;
    }
  }
  void add(int pos, int v) {
    while (pos > 0) {
      val[pos] += v;
      pos -= lowbit(pos);
    }
  }
  int sum(int pos) {
    int res = 0;
    while (pos <= N) {
      res += val[pos];
      pos += lowbit(pos);
    }
    return res;
  }
} T;
int saveori[maxn];
int saveans[maxn];
struct node {
  int l, r;
  int id;
  bool operator<(const node &n) const {
    if (r != n.r) {
      return r < n.r;
    } else {
      return l < n.l;
    }
  }
} saveq[maxn];
int n, q;
int cjladd(int x, int r) {
  int l = 1;
  int pos = -1;
  while (l <= r) {
    int mid = l + r;
    mid /= 2;
    if (T.sum(mid) >= x) {
      pos = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return pos;
}
void solve() {
  cin >> n >> q;
  T.init(n + 10);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &saveori[i]);
    saveori[i] = i - saveori[i];
  }
  for (int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    saveq[i].l = 1 + x;
    saveq[i].r = n - y;
    saveq[i].id = i;
  }
  sort(saveq + 1, saveq + 1 + q);
  int nowrarr = 1;
  for (int i = 1; i <= q; i++) {
    node nownode = saveq[i];
    while (nowrarr <= nownode.r) {
      if (saveori[nowrarr] == 0) {
        T.add(nowrarr, 1);
        nowrarr++;
      } else if (saveori[nowrarr] < 0) {
        nowrarr++;
        continue;
      } else {
        int d = cjladd(saveori[nowrarr], nowrarr);
        if (d != -1) {
          T.add(d, 1);
        }
        nowrarr++;
      }
    }
    saveans[nownode.id] = T.sum(nownode.l);
  }
  for (int i = 1; i <= q; i++) {
    cout << saveans[i] << endl;
  }
}
int main() { solve(); }
