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 inf = 0x3f3f3f3f;
const int N = 3e5 + 100;
int a[N], ans[N];
vector<pair<int, int>> q[N];
struct Node {
  int l, r, len;
  long long sum, lazy;
} tree[N << 2];
void pushup(int k) { tree[k].sum = tree[k << 1].sum + tree[k << 1 | 1].sum; }
void pushdown(int k) {
  if (tree[k].lazy) {
    long long lz = tree[k].lazy;
    tree[k].lazy = 0;
    tree[k << 1].sum += tree[k << 1].len * lz;
    tree[k << 1 | 1].sum += tree[k << 1 | 1].len * lz;
    tree[k << 1].lazy += lz;
    tree[k << 1 | 1].lazy += lz;
  }
}
void build(int k, int l, int r) {
  tree[k].l = l;
  tree[k].r = r;
  tree[k].len = r - l + 1;
  tree[k].sum = tree[k].lazy = 0;
  if (l == r) return;
  int mid = l + r >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  pushup(k);
}
void update(int k, int l, int r) {
  if (l > r) return;
  if (tree[k].l > r || tree[k].r < l) return;
  if (tree[k].l >= l && tree[k].r <= r) {
    tree[k].sum += tree[k].len;
    tree[k].lazy++;
    return;
  }
  pushdown(k);
  update(k << 1, l, r);
  update(k << 1 | 1, l, r);
  pushup(k);
}
long long query(int k, int pos) {
  if (tree[k].l == tree[k].r) return tree[k].sum;
  pushdown(k);
  int mid = tree[k].l + tree[k].r >> 1;
  if (pos <= mid)
    return query(k << 1, pos);
  else
    return query(k << 1 | 1, pos);
}
int get_pos(int i) {
  int l = 1, r = i, ans = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (query(1, mid) >= a[i]) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i] = i - a[i];
    if (a[i] < 0) a[i] = inf;
  }
  for (int i = 1; i <= m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l++;
    r = n - r;
    q[r].emplace_back(l, i);
  }
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    int pos = get_pos(i);
    update(1, 1, pos);
    for (int j = 0; j < q[i].size(); j++)
      ans[q[i][j].second] = query(1, q[i][j].first);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
