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 MAX_N = 2e6 + 10;
const long long mod = 1e9 + 7;
const long long INF = 1e9;
inline long long bpow(long long t, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * t) % mod;
    t = (t * t) % mod, n >>= 1;
  }
  return ans;
}
long long out[MAX_N];
int arr[MAX_N];
int sq;
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if ((a.first.first / sq) != (b.first.first / sq))
    return (a.first.first / sq) < (b.first.first / sq);
  return a.first.second < b.first.second;
}
int main() {
  int n, t, k;
  scanf("%d%d", &n, &t);
  sq = sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  vector<pair<pair<int, int>, int>> que;
  for (int i = 1; i <= t; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    que.push_back({{a, b}, i});
  }
  sort(begin(que), end(que), cmp);
  int l = 1, r = 1;
  unordered_map<int, int> cnt;
  cnt[arr[l]]++;
  long long ans = (arr[l] == cnt[arr[l]]);
  for (auto inp : que) {
    int a = inp.first.first, b = inp.first.second;
    if (a == l && r == b) {
      out[inp.second] = ans;
      continue;
    }
    while (r < b) {
      r++;
      if (cnt[arr[r]] == arr[r])
        cnt[arr[r]]++, ans--;
      else if ((cnt[arr[r]] + 1) == arr[r])
        cnt[arr[r]]++, ans++;
      else
        cnt[arr[r]]++;
    }
    while (r > b) {
      if (cnt[arr[r]] == arr[r])
        cnt[arr[r]]--, ans--;
      else if ((cnt[arr[r]] - 1) == arr[r])
        cnt[arr[r]]--, ans++;
      else
        cnt[arr[r]]--;
      r--;
    }
    while (l < a) {
      if (cnt[arr[l]] == arr[l])
        cnt[arr[l]]--, ans--;
      else if ((cnt[arr[l]] - 1) == arr[l])
        cnt[arr[l]]--, ans++;
      else
        cnt[arr[l]]--;
      l++;
    }
    while (l > a) {
      l--;
      if (cnt[arr[l]] == arr[l])
        cnt[arr[l]]++, ans--;
      else if ((cnt[arr[l]] + 1) == arr[l])
        cnt[arr[l]]++, ans++;
      else
        cnt[arr[l]]++;
    }
    out[inp.second] = ans;
  }
  for (long long i = 1; i <= t; i++) printf("%lld\n", out[i]);
}
