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 MOD = 1e9 + 7;
const long long N = 3e5 + 5;
long long seg[4 * N];
long long get(long long l, long long u, long long i, long long k) {
  if (l == u) {
    if (seg[i] >= k) return u;
    return -1;
  }
  if (seg[(i * 2 + 2)] >= k) return get(((l + u) / 2) + 1, u, (i * 2 + 2), k);
  return get(l, ((l + u) / 2), (i * 2 + 1), k - seg[(i * 2 + 2)]);
}
long long update(long long l, long long u, long long i, long long x) {
  if (l >= x && u <= x) return (++seg[i]);
  if (l > x || u < x) return seg[i];
  return seg[i] = update(l, ((l + u) / 2), (i * 2 + 1), x) +
                  update(((l + u) / 2) + 1, u, (i * 2 + 2), x);
}
long long query(long long l, long long u, long long i, long long ll,
                long long uu) {
  if (l >= ll && u <= uu) return seg[i];
  if (l > uu || u < ll) return 0;
  return query(l, ((l + u) / 2), (i * 2 + 1), ll, uu) +
         query(((l + u) / 2) + 1, u, (i * 2 + 2), ll, uu);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, q;
  cin >> n >> q;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<long long> dis[n];
  for (long long i = 0; i < n; i++) {
    if ((i - a[i] + 1) < 0) {
      continue;
    }
    long long val = min(i, get(0, n - 1, 0, (i - a[i] + 1)));
    if (val < 0) continue;
    update(0, n - 1, 0, val);
    dis[val].push_back(i);
  }
  for (long long i = 0; i < (4 * N); i++) {
    seg[i] = 0;
  }
  vector<pair<long long, long long>> qq[n];
  long long ans[q];
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    qq[x].push_back({n - 1 - y, i});
  }
  for (long long i = n - 1; i >= 0; i--) {
    for (long long j : dis[i]) {
      update(0, n - 1, 0, j);
    }
    for (pair<long long, long long> j : qq[i]) {
      ans[j.second] = query(0, n - 1, 0, i, j.first);
    }
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << endl;
}
