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 = 3e5 + 5;
const long long mod = 998244353;
const long long INF = 1e18 + 5;
long long n, q;
long long a[N];
long long bit[N];
void update(long long idx, long long delta) {
  for (; idx < N; idx += idx & -idx) bit[idx] += delta;
}
long long sum(long long id) {
  long long ret = 0;
  for (long long i = id; i > 0; i -= i & -i) ret += bit[i];
  return ret;
}
void solve() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  vector<long long> ans(q), left(q);
  vector<vector<long long>> endshere(n + 1);
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    long long l = x + 1, r = n - y;
    left[i] = l;
    endshere[r].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      long long l = 0, r = i;
      while (l < r) {
        long long mid = (l + r + 1) / 2;
        if (sum(mid) >= a[i])
          l = mid;
        else
          r = mid - 1;
      }
      update(1, 1);
      update(r + 1, -1);
    }
    for (auto k : endshere[i]) ans[k] = sum(left[k]);
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
