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;
int n, q;
struct query {
  int pos, l;
};
struct Fenwick {
  int n;
  vector<int> aib;
  Fenwick(int _n) {
    aib.resize(_n + 1);
    n = _n;
  }
  void upd(int pos, int val) {
    for (int i = pos; i <= n; i += (i & -i)) {
      aib[i] += val;
    }
  }
  void rupd(int l, int r) {
    if (l > r) return;
    upd(l, 1);
    if (r <= n) upd(r + 1, -1);
  }
  int qry(int x) {
    int sol = 0;
    for (; x; x -= (x & -x)) sol += aib[x];
    return sol;
  }
};
int main() {
  cin >> n >> q;
  vector<int> a(n + 1);
  Fenwick v(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  vector<query> s[n + 1];
  vector<int> ans(q + 1);
  for (int i = 1; i <= q; ++i) {
    int x, y;
    cin >> x >> y;
    int l = x + 1, r = n - y;
    s[r].push_back({i, l});
  }
  for (int r = 1; r <= n; ++r) {
    if (a[r] >= 0) {
      int pas = (1 << 20), pos = 0;
      while (pas) {
        do {
          pas >>= 1;
        } while (pos + pas > r && pas);
        if (v.qry(pas + pos) >= a[r]) pos += pas;
      }
      v.rupd(1, pos);
    }
    for (auto it : s[r]) {
      ans[it.pos] = v.qry(it.l);
    }
  }
  for (int i = 1; i <= q; ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}
