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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
int tr[500005 * 3], tr1[500005 * 3], laz[500005 * 3], a[500005], b[500005];
void make_update(long long n, long long b, long long e) {
  long long mid = (b + e) >> 1, l = n * 2, r = l + 1;
  long long x = laz[n];
  tr[l] -= x;
  tr[r] -= x;
  laz[l] += x;
  laz[r] += x;
  laz[n] = 0;
}
void update(long long n, long long b, long long e, long long x, long long val) {
  if (b == e) {
    tr[n] = val;
    tr1[n] = b;
    return;
  }
  long long mid = (b + e) >> 1, l = n * 2, r = l + 1;
  make_update(n, b, e);
  if (x <= mid)
    update(l, b, mid, x, val);
  else
    update(r, mid + 1, e, x, val);
  tr[n] = min(tr[l], tr[r]);
  if (tr[n] == tr[r])
    tr1[n] = tr1[r];
  else
    tr1[n] = tr1[l];
}
void update1(long long n, long long b, long long e, long long x, long long y) {
  if (b > y || e < x) return;
  if (b >= x && e <= y) {
    tr[n]--;
    laz[n]++;
    return;
  }
  long long mid = (b + e) >> 1, l = n * 2, r = l + 1;
  make_update(n, b, e);
  update1(l, b, mid, x, y);
  update1(r, mid + 1, e, x, y);
  tr[n] = min(tr[l], tr[r]);
  if (tr[n] == tr[r])
    tr1[n] = tr1[r];
  else
    tr1[n] = tr1[l];
}
long long t[500005];
void update(long long x, long long val) {
  while (x < 500005) {
    t[x] += val;
    x += x & (-x);
  }
}
long long queury(long long x) {
  long long s = 0;
  while (x > 0) {
    s += t[x];
    x -= x & (-x);
  }
  return s;
}
vector<int> adj[500005];
vector<pair<int, int> > adj1[500005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q, i, j;
  while (cin >> n >> q) {
    for (i = 1; i <= n; i++) cin >> a[i];
    priority_queue<long long> pq;
    for (i = 1; i <= n; i++) {
      if (a[i] == i) {
        b[i] = i;
        pq.push(i);
        update(1, 1, n, i, 123456789);
      } else if (i > a[i]) {
        update(1, 1, n, i, i - a[i]);
      } else
        update(1, 1, n, i, 123456789);
    }
    while (pq.size() > 0) {
      long long r = pq.top();
      pq.pop();
      update1(1, 1, n, r + 1, n);
      while (1) {
        if (tr[1] != 0) break;
        pq.push(tr1[1]);
        b[tr1[1]] = b[r];
        update(1, 1, n, tr1[1], 123456789);
      }
    }
    for (i = 1; i <= n; i++) {
      if (b[i] == 0) continue;
      adj[i].push_back(b[i]);
    }
    for (i = 1; i <= q; i++) {
      int u, v;
      cin >> u >> v;
      adj1[n - v].push_back(make_pair(u + 1, i));
    }
    for (i = 1; i <= n; i++) {
      for (auto j : adj[i]) {
        update(j, 1);
      }
      for (auto j : adj1[i]) {
        long long ind = j.second;
        long long x = j.first;
        b[ind] = queury(i) - queury(x - 1);
      }
    }
    for (i = 1; i <= q; i++) cout << b[i] << endl;
  }
  return 0;
}
