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 long long N = 3 * 1e5 + 5;
long long n, q;
vector<pair<long long, long long>> query[N];
struct BIT {
  long long arr[N];
  void update(long long i, long long x) {
    while (i <= n) {
      arr[i] += x;
      i += i & -i;
    }
  }
  long long get(long long i) {
    long long ans = 0;
    while (i >= 1) {
      ans += arr[i];
      i -= i & -i;
    }
    return ans;
  }
} bit;
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = i - a[i];
    if (a[i] < 0) {
      a[i] = 1e9;
    }
  }
  for (long long i = 1; i <= q; i++) {
    long long x, y;
    cin >> x >> y;
    x = x + 1;
    y = n - y;
    query[y].push_back({x, i});
  }
  vector<long long> ans(q + 1);
  for (long long r = 1; r <= n; r++) {
    long long left = 0, right = r + 1;
    while (left + 1 < right) {
      long long mid = left + (right - left) / 2;
      if (bit.get(mid) >= a[r]) {
        left = mid;
      } else {
        right = mid;
      }
    }
    if (left != 0) {
      bit.update(1, 1);
      bit.update(left + 1, -1);
    }
    for (auto p : query[r]) {
      ans[p.second] = bit.get(p.first);
    }
  }
  for (long long i = 1; i <= q; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}
