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;
int row, col, n, m, q, cnt, ara[100000 + 5], ans[100000 + 5];
unordered_map<int, int> mark;
void add(int pos) {
  if (mark[ara[pos]] == ara[pos]) --cnt;
  mark[ara[pos]]++;
  if (mark[ara[pos]] == ara[pos]) {
    cnt++;
  }
}
void del(int pos) {
  if (mark[ara[pos]] == ara[pos]) --cnt;
  mark[ara[pos]] -= 1;
  if (mark[ara[pos]] == ara[pos]) ++cnt;
}
int main() {
  int T;
  T = 1;
  mark.reserve(64);
  mark.max_load_factor(0.25);
  for (int tc = 1; tc <= T; tc++) {
    vector<pair<pair<int, int>, pair<int, int> > > v;
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++) scanf("%d", &ara[i]);
    int sq = sqrt(n);
    for (int i = 1; i <= q; i++) {
      int lft, rht;
      scanf("%d %d", &lft, &rht);
      v.push_back({{lft / sq, rht}, {lft, i}});
    }
    sort(v.begin(), v.end());
    ;
    ara[0] = ara[1];
    ara[n + 1] = ara[n];
    int lft = 1, rht = 0;
    for (pair<pair<int, int>, pair<int, int> > it : v) {
      int queryLft = it.second.first, queryRht = it.first.second;
      while (queryRht > rht) add(++rht);
      while (queryRht < rht) del(rht--);
      while (queryLft > lft) del(lft++);
      while (queryLft < lft) add(--lft);
      int idx = it.second.second;
      ans[idx] = cnt;
    }
    for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  }
}
