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 + 10;
long long a[maxn];
long long s[maxn];
long long n, q;
int rr;
long long query(long long k) {
  long long ans = 0;
  while (k <= n && k > 0) {
    ans += s[k];
    k += k & (-k);
  }
  return ans;
}
void add(long long k) {
  while (k > 0 && k < maxn) {
    s[k]++;
    k -= k & (-k);
  }
}
long long find(long long x) {
  long long ans = -1;
  long long l = 1, r = rr;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (query(mid) >= x) {
      ans = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return ans;
}
vector<pair<int, int>> r[maxn];
int A[maxn];
int main(void) {
  scanf("%lld%lld", &n, &q);
  for (int i = 0; i <= n; i++) s[i] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] = i - a[i];
  }
  for (int i = 1, x, y; i <= q; i++) {
    scanf("%d%d", &x, &y);
    r[n - y].emplace_back(make_pair(1 + x, i));
  }
  for (int i = 1; i <= n; i++) {
    rr = i;
    long long k = find(a[i]);
    if (k > i) k = i;
    if (k != -1 && a[i] >= 0) add(k);
    for (auto x : r[i]) {
      A[x.second] = query(x.first);
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", A[i]);
  return 0;
}
