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 = 100500;
int a[maxn];
int n;
map<int, int> indexes;
struct Triple {
  int first;
  int second;
  int third;
};
vector<Triple> segments[maxn];
queue<int> queues[maxn];
map<int, int> backmap;
vector<int> valids;
int main() {
  int m;
  scanf("%d", &n);
  scanf("%d", &m);
  int i;
  for (i = 0; i < n; ++i) scanf("%d", &a[i]);
  int ind;
  int curi = 1;
  for (i = 0; i < n; ++i) {
    if (indexes[a[i]] == 0) {
      indexes[a[i]] = curi;
      backmap[curi] = a[i];
      ++curi;
    }
    ind = indexes[a[i]];
    if (queues[ind].size() == a[i]) {
      Triple tri;
      tri.first = queues[ind].front();
      tri.second = queues[ind].back();
      tri.third = i;
      segments[ind].push_back(tri);
      queues[ind].pop();
    }
    queues[ind].push(i);
  }
  for (i = 1; i < curi; ++i)
    if (queues[i].size() == backmap[i]) {
      Triple tri;
      tri.first = queues[i].front();
      tri.second = queues[i].back();
      tri.third = n;
      segments[i].push_back(tri);
    }
  for (i = 1; i < curi; ++i)
    if (!segments[i].empty()) valids.push_back(i);
  int j;
  int l, r;
  int bsl, bsr, bsm;
  for (j = 0; j < m; ++j) {
    scanf("%d", &l), scanf("%d", &r);
    --l;
    --r;
    int ans = 0;
    for (i = 0; i < valids.size(); ++i) {
      ind = valids[i];
      bsl = 0;
      bsr = segments[ind].size() - 1;
      while (bsl < bsr) {
        bsm = (bsl + bsr) / 2;
        if (l > segments[ind][bsm].first)
          bsl = bsm + 1;
        else
          bsr = bsm;
      }
      if (l <= segments[ind][bsr].first && r >= segments[ind][bsr].second &&
          r < segments[ind][bsr].third)
        ++ans;
    }
    printf("%d\n", ans);
    ;
  }
  return 0;
}
