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 N = 100005, Q = N, M = N;
int n, q, a[N], b[N], qi[Q], ql[Q], qr[Q], qa[Q], cnt = 0, l = 0, r = -1,
                                                  freq[M];
pair<int, int> p[N];
void add(int val) {
  cnt -= freq[val] == b[val];
  freq[val]++;
  cnt += freq[val] == b[val];
}
void rem(int val) {
  cnt -= freq[val] == b[val];
  freq[val]--;
  cnt += freq[val] == b[val];
}
void update(int cur) {
  while (r < qr[cur]) add(a[++r]);
  while (l > ql[cur]) add(a[--l]);
  while (r > qr[cur]) rem(a[r--]);
  while (l < ql[cur]) rem(a[l++]);
}
void mo() {
  int sq = sqrt(n);
  sort(qi, qi + q, [sq](int a, int b) {
    return make_pair(ql[a] / sq, qr[a]) < make_pair(ql[b] / sq, qr[b]);
  });
  for (int i = 0; i < q; i++) {
    int cur = qi[i];
    update(cur);
    qa[cur] = cnt;
  }
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    p[i] = {a[i], i};
  }
  sort(p, p + n);
  int sz = 1;
  a[p[0].second] = sz;
  b[sz] = p[0].first;
  for (int i = 1; i < n; i++) {
    if (p[i].first != p[i - 1].first) {
      sz++;
    }
    a[p[i].second] = sz;
    b[sz] = p[i].first;
  }
  for (int i = 0; i < q; i++) {
    qi[i] = i;
    scanf("%d %d", &ql[i], &qr[i]);
    ql[i]--;
    qr[i]--;
  }
  mo();
  for (int i = 0; i < q; i++) {
    printf("%d\n", qa[i]);
  }
  return 0;
}
