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>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, q;
  std::cin >> n >> q;
  std::vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    --a[i];
    if (a[i] > i) a[i] = -1;
    a[i] = i - a[i];
  }
  std::vector<std::vector<int>> dp(4 * n);
  std::function<void(int, int, int)> solve = [&](int p, int l, int r) {
    dp[p].resize(r - l + 1);
    if (r - l == 1) {
      dp[p][1] = a[l];
    } else {
      int m = (l + r) / 2;
      solve(2 * p, l, m);
      solve(2 * p + 1, m, r);
      for (int i = 0, j = 0, k = 0, s = 0; i <= r - l; ++i) {
        while (j + k < i) {
          if (k + 1 <= r - m && dp[2 * p + 1][k + 1] <= s + j) {
            ++k;
          } else if (j + 1 <= m - l && dp[2 * p][j + 1] <= s) {
            ++j;
          } else {
            s = std::min(j + 1 <= m - l ? dp[2 * p][j + 1] : n + 1,
                         k + 1 <= r - m ? dp[2 * p + 1][k + 1] - j : n + 1);
          }
        }
        dp[p][i] = s;
      }
    }
  };
  solve(1, 0, n);
  std::function<int(int, int, int, int, int, int)> query =
      [&](int p, int l, int r, int x, int y, int s) -> int {
    if (l >= y || r <= x) {
      return s;
    } else if (l >= x && r <= y) {
      return s + std::upper_bound(dp[p].begin(), dp[p].end(), s) -
             dp[p].begin() - 1;
    } else {
      int m = (l + r) / 2;
      s = query(2 * p, l, m, x, y, s);
      return query(2 * p + 1, m, r, x, y, s);
    }
  };
  while (q--) {
    int l, r;
    std::cin >> l >> r;
    r = n - r;
    std::cout << query(1, 0, n, l, r, 0) << "\n";
  }
  return 0;
}
