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>
#pragma comment(linker, "/STACK:100000000,100000000")
using namespace std;
const long long inf = 1e9 + 7;
const long long mod = 1e9 + 7;
const long double eps = 1e-10;
const long double PI = 2 * acos(0.0);
const long double E = 2.71828;
const long long seed = 13331;
pair<pair<long long, long long>, long long> p[111111];
long long cnt[111111], a[1111111];
long long n, cur = 0;
long long ans[111111];
bool cmp(pair<pair<long long, long long>, long long> a,
         pair<pair<long long, long long>, long long> b) {
  if (a.first.first / 300 != b.first.first / 300)
    return a.first.first / 300 < b.first.first / 300;
  return a.first.second < b.first.second;
}
void add(int x) {
  if (x > n) return;
  if (cnt[x] == x) cur--;
  cnt[x]++;
  if (cnt[x] == x) cur++;
}
void del(int x) {
  if (x > n) return;
  if (cnt[x] == x) cur--;
  cnt[x]--;
  if (cnt[x] == x) cur++;
}
int main(void) {
  int m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++)
    cin >> p[i].first.first >> p[i].first.second, p[i].second = i;
  sort(p, p + m, cmp);
  int aa = 0, bb = 0;
  if (a[0] <= n) cnt[a[0]]++;
  if (a[0] == 1) cur++;
  for (int i = 0; i < m; i++) {
    int l = p[i].first.first - 1, r = p[i].first.second - 1, to = p[i].second;
    while (bb < r) bb++, add(a[bb]);
    while (bb > r) bb--, del(a[bb + 1]);
    while (aa < l) aa++, del(a[aa - 1]);
    while (aa > l) aa--, add(a[aa]);
    ans[to] = cur;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}
