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;
vector<pair<int, int> > vv[300005];
int tree[300005 * 4];
int ans[300005];
int arr[300005];
void update(int node, int b, int e, int i) {
  if (b > i || e < i) return;
  if (b == e && b == i) {
    tree[node]++;
    return;
  }
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  update(left, b, mid, i);
  update(right, mid + 1, e, i);
  tree[node] = tree[left] + tree[right];
}
int query1(int node, int b, int e, int v) {
  if (b == e) return b;
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  int ret;
  if (tree[right] >= v)
    ret = query1(right, mid + 1, e, v);
  else
    ret = query1(left, b, mid, v - tree[right]);
  return ret;
}
int query(int node, int b, int e, int l, int r) {
  if (l > r) return 0;
  if (b > r || e < l) return 0;
  if (b >= l && e <= r) return tree[node];
  int left = 2 * node, right = left + 1, mid = (b + e) / 2;
  int p = query(left, b, mid, l, r);
  int q = query(right, mid + 1, e, l, r);
  int ret = p + q;
  return ret;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  for (int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    vv[n - y].push_back(make_pair(x + 1, i));
  }
  for (int cs = 1; cs <= n; cs++) {
    if (cs - arr[cs] == 0)
      update(1, 1, n, cs);
    else if (cs - arr[cs] > 0 && cs - arr[cs] <= tree[1]) {
      int id = query1(1, 1, n, cs - arr[cs]);
      update(1, 1, n, id);
    }
    for (int i = 0; i < vv[cs].size(); i++) {
      ans[vv[cs][i].second] = query(1, 1, n, vv[cs][i].first, cs);
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
