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 N = 3e5 + 5;
int n, m, a[N], qL[N], qR[N], bit[N], ans[N];
vector<int> vec[N];
int lowbit(int x) { return x & (-x); }
void insert(int k, int x) {
  for (; k <= n; k += lowbit(k)) bit[k] += x;
  return;
}
int query(int k) {
  int res = 0;
  for (; k > 0; k -= lowbit(k)) res += bit[k];
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    qL[i] = x + 1, qR[i] = n - y;
    vec[qR[i]].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    if (i >= a[i]) {
      int c = i - a[i], l = 1, r = i, nR = 0;
      while (l <= r) {
        int mid = l + r >> 1;
        if (query(mid) >= c)
          nR = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      if (nR > 0) insert(1, 1), insert(nR + 1, -1);
    }
    for (int j = 0, sze = vec[i].size(); j < sze; ++j) {
      int k = vec[i][j];
      ans[k] = query(qL[k]);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}
