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>
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
  putchar(' ');
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
const long double Pi = acos(-1);
using namespace std;
int n, q, im[300005], it[300005 << 2], a[300005];
vector<int> t[300005 << 2];
void Build(int l, int r, int id) {
  if (l == r) {
    t[id].push_back(im[l]);
    return;
  }
  int mid = (l + r) >> 1;
  Build(l, mid, 2 * id);
  Build(mid + 1, r, 2 * id + 1);
  t[id].resize(r - l + 1);
  merge(t[2 * id].begin(), t[2 * id].end(), t[2 * id + 1].begin(),
        t[2 * id + 1].end(), t[id].begin());
}
int Get(int l, int r, int id, int u, int v) {
  if (l > v || r < u) return 0;
  if (u <= l && r <= v) {
    return upper_bound(t[id].begin(), t[id].end(), v) -
           lower_bound(t[id].begin(), t[id].end(), u);
  }
  int mid = (l + r) >> 1;
  return Get(l, mid, 2 * id, u, v) + Get(mid + 1, r, 2 * id + 1, u, v);
}
void Updateit(int l, int r, int id, int x) {
  ++it[id];
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (x <= mid)
    Updateit(l, mid, 2 * id, x);
  else
    Updateit(mid + 1, r, 2 * id + 1, x);
}
int Getit(int l, int r, int id, int u, int v) {
  if (l > v || r < u) return 0;
  if (u <= l && r <= v) return it[id];
  int mid = (l + r) >> 1;
  return Getit(l, mid, 2 * id, u, v) + Getit(mid + 1, r, 2 * id + 1, u, v);
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie();
  if (fopen(".inp", "r")) freopen(".inp", "r", stdin);
  cin >> n >> q;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    a[i] -= i;
    if (a[i] > 0) continue;
    int ll = 1, rr = i + 1;
    while (rr - ll > 1) {
      int mid = (ll + rr) >> 1;
      if (Getit(1, n, 1, mid, i) >= -a[i])
        ll = mid;
      else
        rr = mid;
    }
    if (Getit(1, n, 1, ll, i) >= -a[i]) {
      im[i] = ll;
      Updateit(1, n, 1, ll);
    }
  }
  Build(1, n, 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    ++l;
    r = n - r;
    cout << Get(1, n, 1, l, r) << '\n';
  }
  return 0;
}
