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 N = 100010, MOD = 1000000007;
set<int> ste;
struct node {
  int l, r, id;
} g[N];
int a[N], b[N], c[N], res[N], num[N];
int n, m, tmp, unit;
void add(int i) {
  if (num[a[i]] == c[i] - 1)
    tmp++;
  else if (num[a[i]] == c[i])
    tmp--;
  num[a[i]]++;
}
void del(int i) {
  if (num[a[i]] == c[i])
    tmp--;
  else if (num[a[i]] == c[i] + 1)
    tmp++;
  num[a[i]]--;
}
void solve() {
  unit = (int)sqrt(1.0 * n);
  memset(num, 0, sizeof num);
  sort(g + 1, g + 1 + m, [](node a, node b) {
    return a.l / unit != b.l / unit ? a.l / unit < b.l / unit : a.r < b.r;
  });
  int l = 1, r = 0;
  tmp = 0;
  for (int i = 1; i <= m; i++) {
    while (r < g[i].r) add(++r);
    while (r > g[i].r) del(r--);
    while (l < g[i].l) del(l++);
    while (l > g[i].l) add(--l);
    res[g[i].id] = tmp;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), c[i] = b[i] = a[i];
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
  for (int i = 1; i <= m; i++) scanf("%d%d", &g[i].l, &g[i].r), g[i].id = i;
  solve();
  return 0;
}
