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;
vector<pair<long long, long long> > v, q;
map<pair<long long, long long>, long long> mp;
long long n, m, i, x, y, b[100005], a[100005], ans = 0, l, r, j, s;
void add(long long x) {
  if (x > 1e5) return;
  if (b[x] == x) ans--;
  b[x]++;
  if (b[x] == x) ans++;
}
void rem(long long x) {
  if (x > 1e5) return;
  if (b[x] == x) ans--;
  b[x]--;
  if (b[x] == x) ans++;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first / s == b.first / s) return a.second < b.second;
  return a.first / s < b.first / s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  s = sqrt(n);
  for (i = 0; i < n; ++i) cin >> a[i];
  for (i = 0; i < m; ++i) {
    cin >> x >> y;
    x--;
    y--;
    v.push_back({x, y});
  }
  q = v;
  sort(v.begin(), v.end(), cmp);
  l = 0;
  r = 0;
  for (i = 0; i < m; ++i) {
    while (l < v[i].first) {
      rem(a[l]);
      l++;
    }
    while (l > v[i].first) {
      add(a[l - 1]);
      l--;
    }
    while (r <= v[i].second) {
      add(a[r]);
      r++;
    }
    while (r > v[i].second + 1) {
      rem(a[r - 1]);
      r--;
    }
    mp[{l, r - 1}] = ans;
  }
  for (i = 0; i < m; ++i) cout << mp[q[i]] << endl;
  return 0;
}
