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 = 1e5 + 5, SQRT = 460;
int n, m, res;
int ans[MAXN];
int Arr[MAXN];
int Arr2[MAXN];
int Arr3[MAXN];
int cnt[MAXN];
int IND = 1;
unordered_map<int, int> D1, D2;
struct query {
  int id, l, r;
} q[MAXN];
void Add(int ind) {
  int tmp = Arr2[ind];
  if (cnt[tmp] == Arr3[tmp]) res--;
  cnt[tmp]++;
  if (cnt[tmp] == Arr3[tmp]) res++;
}
void Dil(int ind) {
  int tmp = Arr2[ind];
  if (cnt[tmp] == Arr3[tmp]) res--;
  cnt[tmp]--;
  if (cnt[tmp] == Arr3[tmp]) res++;
}
bool cmp(query a, query b) {
  if (a.l / SQRT == b.l / SQRT) return a.r < b.r;
  return a.l < b.l;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &Arr[i]);
    if (D1[Arr[i]] == 0) {
      D1[Arr[i]] = IND;
      D2[IND++] = Arr[i];
    }
    Arr2[i] = D1[Arr[i]];
    Arr3[Arr2[i]] = Arr[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].l--;
    q[i].id = i;
  }
  sort(q, q + m, cmp);
  int L = 0, R = 0;
  for (int i = 0; i < m; i++) {
    while (R < q[i].r) Add(R++);
    while (L > q[i].l) Add(--L);
    while (R > q[i].r) Dil(--R);
    while (L < q[i].l) Dil(L++);
    ans[q[i].id] = res;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
