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 MOD = 1e9 + 7;
const int MAXN = 3e5 + 5;
int n, q;
int fenwick[MAXN];
int lsb(int x) { return x & -x; }
void update(int p, int v) {
  while (p <= n + 2) {
    fenwick[p] += v;
    p += lsb(p);
  }
}
int query(int p) {
  int sum = 0;
  while (p > 0) {
    sum += fenwick[p];
    p -= lsb(p);
  }
  return sum;
}
int main() {
  cin >> n >> q;
  int ar[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    ar[i] -= i;
  }
  priority_queue<pair<pair<int, int>, int>> queries;
  for (int i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    queries.push(make_pair(make_pair(y, x), i));
  }
  int pos = 1;
  int ans[n + 1];
  int out[q + 1];
  int cnt = 0;
  int cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    if (ar[i] <= 0) {
      if (ar[i] + cnt1 >= 0) cnt1++;
    }
    ans[i] = cnt1;
  }
  while (queries.size() > 0 && pos <= n - queries.top().first.first) {
    int l = 0;
    int r = pos;
    if (ar[pos] > 0) goto skip;
    while (r - l > 1) {
      int m = l + (r - l) / 2;
      if (cnt - query(m) + ar[pos] >= 0)
        l = m;
      else
        r = m;
    }
    if (l == 0 && pos == 1 && ans[1] + ar[1] >= 0) {
      update(2, 1);
      cnt++;
      goto skip;
    }
    if (l == 0) goto skip;
    if (cnt - query(r) + ar[pos] >= 0)
      update(r + 1, 1), cnt++;
    else
      update(l + 1, 1), cnt++;
  skip:;
    pos++;
    while (queries.size() > 0 && pos > n - queries.top().first.first) {
      out[queries.top().second] = cnt - query(queries.top().first.second + 1);
      queries.pop();
    }
  }
  for (int i = 1; i <= q; i++) {
    cout << out[i] << "\n";
  }
}
