The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai. 

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4


Output

3
1
#include <bits/stdc++.h>
using namespace std;
struct o {
  int l, r, i;
};
const int N = 100005;
o q[N];
int n, m, a[N], re[N], dd[N], kq, l = 1, r = 1, t;
bool cmp(o c, o d) {
  if (((c.l - 1) / t + 1) != ((d.l - 1) / t + 1))
    return (((c.l - 1) / t + 1) < ((d.l - 1) / t + 1));
  else
    return c.r < d.r;
}
void add(int i) {
  if (i > n) return;
  ++dd[i];
  if (dd[i] == i)
    kq++;
  else if (dd[i] == i + 1)
    kq--;
}
void rev(int i) {
  if (i > n) return;
  --dd[i];
  if (dd[i] == i)
    ++kq;
  else if (dd[i] == i - 1)
    --kq;
}
int main() {
  cin >> n >> m;
  t = ceil(sqrt(n));
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; ++i) {
    cin >> q[i].l >> q[i].r;
    q[i].i = i;
  }
  sort(q + 1, q + m + 1, cmp);
  add(a[1]);
  for (int i = 1; i <= m; ++i) {
    int u = q[i].l, v = q[i].r;
    while (l < u) rev(a[l++]);
    while (l > u) add(a[--l]);
    while (r < v) add(a[++r]);
    while (r > v) rev(a[r--]);
    re[q[i].i] = kq;
  }
  for (int i = 1; i <= m; ++i) cout << re[i] << "\n";
}
