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 = 3e5 + 7;
struct query {
  int l, r, id;
  query(int li, int ri, int idi) {
    l = li;
    r = ri;
    id = idi;
  }
  query() {}
};
bool comp(query q1, query q2) { return q1.r < q2.r; }
int n, q, fenwick[MAXN];
int qry(int idx) {
  int ret = 0;
  for (; idx > 0; idx -= (idx & (-idx))) ret += fenwick[idx];
  return ret;
}
void add(int idx, int val) {
  for (; idx < MAXN; idx += (idx & (-idx))) fenwick[idx] += val;
}
int main() {
  scanf("%d %d", &n, &q);
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = i - a[i];
  }
  vector<query> v;
  vector<int> ans(q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    int li = x + 1;
    int ri = n - y;
    v.push_back(query(li, ri, i));
  }
  sort(v.begin(), v.end(), comp);
  int cur_r = 0;
  for (query qi : v) {
    while (cur_r < qi.r) {
      cur_r++;
      if (a[cur_r] < 0) continue;
      if (a[cur_r] == 0) {
        add(1, +1);
        add(cur_r + 1, -1);
        continue;
      }
      int tst = qry(1);
      if (tst < a[cur_r]) {
        continue;
      }
      int bl = 1;
      int br = cur_r;
      while (bl != br) {
        int mid = (bl + br) / 2;
        int aux = qry(mid);
        if (a[cur_r] > aux)
          br = mid;
        else
          bl = mid + 1;
      }
      add(1, +1);
      add(bl, -1);
    }
    ans[qi.id] = qry(qi.l);
  }
  for (int i : ans) {
    printf("%d\n", i);
  }
}
