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;
using pii = pair<int, int>;
int n, q;
int in[300010];
vector<pii> query[300010];
int ans[300010], out[300010];
struct BITree {
  int num[300010];
  BITree() { memset(num, 0, sizeof(num)); }
  void add(int x, int y) {
    while (x <= n) {
      num[x] += y;
      x += x & -x;
    }
  }
  int query(int x) {
    int re = 0;
    while (x) {
      re += num[x];
      x -= x & -x;
    }
    return re;
  }
} bit;
int find(int now) {
  int all = ans[now];
  int fn = in[now];
  int l = 0, r = now;
  while (l + 1 < r) {
    int m = (l + r) / 2;
    if (all - bit.query(m) >= fn)
      l = m;
    else
      r = m;
  }
  return l;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> in[i];
    in[i] = i - in[i];
  }
  for (int i = 0; i < q; i++) {
    int a, b;
    cin >> a >> b;
    query[n - b].emplace_back(pii(a, i));
  }
  for (int i = 1; i <= n; i++) {
    ans[i] = ans[i - 1];
    if (in[i] >= 0 && ans[i - 1] >= in[i]) {
      int last = find(i);
      bit.add(last + 1, 1);
      ans[i]++;
    }
    for (pii &x : query[i]) out[x.second] = ans[i] - bit.query(x.first);
  }
  for (int i = 0; i < q; i++) cout << out[i] << '\n';
  return 0;
}
