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;
const int mxn = 1e5 + 10;
int a[mxn], f[mxn];
struct query {
  int l, r, i;
} q[mxn];
int ans[mxn];
bool cmp(query a, query b) {
  int bl1 = a.l / 370, bl2 = b.l / 370;
  return (bl1 == bl2 ? a.r < b.r : bl1 < bl2);
}
int cnt = 0;
bool vis[mxn];
void add(int x) {
  if (x > 100000) return;
  if (x == 0) return;
  f[x]++;
  if (f[x] == x) cnt++;
  if (f[x] == x + 1) cnt--;
}
void rem(int x) {
  if (x > 100000) return;
  if (x == 0) return;
  f[x]--;
  if (f[x] == x) cnt++;
  if (f[x] == x - 1) cnt--;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    int l, r;
    cin >> l >> r;
    q[i].l = l, q[i].r = r, q[i].i = i;
  }
  sort(q + 1, q + m + 1, cmp);
  int curr = 0, curl = 0;
  for (int i = 1; i <= m; i++) {
    int l = q[i].l, r = q[i].r;
    while (curl < l) rem(a[curl++]);
    while (curl > l) add(a[--curl]);
    while (curr > r) rem(a[curr--]);
    while (curr < r) add(a[++curr]);
    ans[q[i].i] = cnt;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
}
