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 const N = 3e5 + 10;
int const LOG = 20;
using PII = pair<int, int>;
using ll = long long int;
vector<PII> R[N];
ll t[N];
ll Q(int pos) {
  ll ret = 0;
  for (; pos; pos -= (pos & (-pos))) ret += t[pos];
  return ret;
}
void update(int pos, int val) {
  for (; pos < N; pos += (pos & (-pos))) t[pos] += val;
}
int lower(int val, int lim) {
  int cur_ind = 0, tot = 0;
  for (int i = LOG - 1; i >= 0; i--) {
    if (cur_ind + (1 << i) > lim) continue;
    if (t[cur_ind + (1 << i)] + tot >= val) {
      cur_ind |= (1 << i), tot += t[cur_ind];
    }
  }
  return cur_ind;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i], a[i] = i - a[i];
  for (int i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    R[n - y].emplace_back(x + 1, i);
  }
  vector<int> ans(q + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      int right_valid = lower(a[i], i);
      if (right_valid) {
        update(1, 1);
        update(right_valid + 1, -1);
      }
    }
    for (auto p : R[i]) {
      ans[p.second] = Q(p.first);
    }
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}
