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 MAXN = 100010;
int n, q, k;
int s;
int a[MAXN];
int cnt[MAXN];
int cur = 0;
void add(int v) {
  if (v <= n) {
    cnt[v]++;
    if (cnt[v] == v) {
      cur++;
    }
    if (cnt[v] == v + 1) {
      cur--;
    }
  }
}
void del(int v) {
  if (v <= n) {
    cnt[v]--;
    if (cnt[v] == v) {
      cur++;
    }
    if (cnt[v] == v - 1) {
      cur--;
    }
  }
}
struct query {
  int l, r;
  int ind;
} qry[MAXN];
bool cmp(query a, query b) {
  if (a.l / s != b.l / s) return a.l < b.l;
  return a.r < b.r;
}
int ans[MAXN];
int main() {
  scanf("%d%d", &n, &q);
  s = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &qry[i].l, &qry[i].r);
    qry[i].ind = i;
  }
  sort(qry + 1, qry + q + 1, cmp);
  int cl = 1, cr = 0;
  for (int i = 1; i <= q; i++) {
    int l = qry[i].l;
    int r = qry[i].r;
    int id = qry[i].ind;
    while (cl < l) {
      del(a[cl]);
      cl++;
    }
    while (cl > l) {
      cl--;
      add(a[cl]);
    }
    while (cr < r) {
      cr++;
      add(a[cr]);
    }
    while (cr > r) {
      del(a[cr]);
      cr--;
    }
    ans[id] = cur;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
