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 = 300100;
const int INF = 0x3f3f3f3f;
struct seg {
  seg *fe, *fd;
  int meio, l, r, soma, lazy;
  pair<int, int> mix;
  seg(int x, int y) {
    meio = (x + y) / 2;
    l = x, r = y;
    soma = lazy = 0;
    mix = {0, l};
    if (x == y) return;
    fe = new seg(x, meio);
    fd = new seg(meio + 1, y);
  }
  void lazyup() {
    if (lazy == 0) return;
    soma += (r - l + 1) * lazy;
    mix.first += lazy;
    if (l != r) fe->lazy += lazy, fd->lazy += lazy;
    lazy = 0;
  }
  void up(int a, int b, int novo) {
    lazyup();
    if (l >= a && r <= b) {
      lazy += novo;
      lazyup();
      return;
    }
    if (l > b || r < a) return;
    fe->up(a, b, novo), fd->up(a, b, novo);
    if (fe->mix.first <= fd->mix.first)
      mix = fe->mix;
    else
      mix = fd->mix;
    soma = fe->soma + fd->soma;
  }
  pair<int, int> get(int a, int b) {
    lazyup();
    if (l >= a && r <= b) return mix;
    if (l > b || r < a) return {INF, INF};
    pair<int, int> re = fe->get(a, b), rd = fd->get(a, b);
    if (re.first <= rd.first) return re;
    return rd;
  }
  int gets(int a, int b) {
    lazyup();
    if (l >= a && r <= b) return soma;
    if (l > b || r < a) return 0;
    return fe->gets(a, b) + fd->gets(a, b);
  }
};
int v[MAXN];
int resp[MAXN];
vector<pair<int, int>> vec[MAXN];
seg st(0, MAXN);
seg sum(0, MAXN);
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    vec[x].push_back({y, i});
  }
  for (int i = n; i >= 0; i--) {
    for (auto j : vec[i]) {
      resp[j.second] = sum.gets(i + 1, n - j.first);
    }
    if (v[i] <= i)
      st.up(i, i, i - v[i]);
    else
      st.up(i, i, INF);
    while (true) {
      pair<int, int> h = st.get(i, n);
      if (h.first > 0) break;
      st.up(h.second, h.second, INF);
      st.up(h.second + 1, n, -1);
      sum.up(h.second, h.second, 1);
    }
  }
  for (int i = 1; i <= q; i++) cout << resp[i] << endl;
}
