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 = 100010;
int n, m, k;
struct node {
  int l, r, id;
} Q[maxn];
int pos[maxn];
bool cmp(const node &a, const node &b) {
  return pos[a.l] < pos[b.l] ||
         (pos[a.l] == pos[b.l] && (pos[a.l] & 1 ? a.r < b.r : a.r > b.r));
}
long long num[maxn * 2];
long long a[maxn], b[maxn], c[maxn];
long long ans[maxn], ans2[maxn];
int L = 0, R = 0;
long long Ans = 0;
void add(int x) {
  num[c[x]]++;
  if (num[c[x]] == a[x])
    Ans++;
  else if (num[c[x]] == a[x] + 1)
    Ans--;
}
void del(int x) {
  num[c[x]]--;
  if (num[c[x]] == a[x])
    Ans++;
  else if (num[c[x]] + 1 == a[x])
    Ans--;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    memset(num, 0, sizeof(num));
    int sz = sqrt(n * 1.0);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      pos[i] = i / sz;
      b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    int num1 = unique(b + 1, b + n + 1) - b - 1;
    for (int i = 1; i <= n; i++)
      c[i] = lower_bound(b + 1, b + num1 + 1, a[i]) - b;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &Q[i].l, &Q[i].r);
      Q[i].id = i;
    }
    sort(Q + 1, Q + 1 + m, cmp);
    num[0] = 0;
    L = 1;
    R = 0;
    Ans = 0;
    for (int i = 1; i <= m; i++) {
      while (R < Q[i].r) {
        R++;
        add(R);
      }
      while (L > Q[i].l) {
        L--;
        add(L);
      }
      while (L < Q[i].l) {
        del(L);
        L++;
      }
      while (R > Q[i].r) {
        del(R);
        R--;
      }
      ans[Q[i].id] = Ans;
    }
    for (int i = 1; i <= m; i++) printf("%I64d\n", ans[i]);
  }
  return 0;
}
