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 = 100005;
const int MAXD = 450;
int seg[MAXN][MAXD] = {};
int a[MAXN];
map<int, int> mp;
int n, m;
vector<int> can;
map<int, int> ind;
inline void add(short *res, short *a, short *b) {
  for (int i = 0; i < (can.size()); ++i) {
    res[i] = a[i] + b[i];
  }
}
inline void add(short *res, short *a) {
  for (int i = 0; i < (can.size()); ++i) res[i] += a[i];
}
inline int ask(int a, int b) {
  int res = 0;
  for (int i = 0; i < (can.size()); ++i) {
    if (can[i] == seg[b][i] - seg[a - 1][i]) {
      res++;
    }
  }
  return res;
}
int main(void) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= (n); ++i) scanf("%d", a + i), ++mp[a[i]];
  for (__typeof((mp).begin()) it = (mp).begin(); it != (mp).end(); ++it)
    if (it->first <= it->second) {
      ind[it->first] = can.size();
      can.push_back(it->first);
    }
  for (int i = 1; i <= (n); ++i) {
    for (int j = 0; j < (can.size()); ++j) seg[i][j] = seg[i - 1][j];
    if (ind.count(a[i])) seg[i][ind[a[i]]]++;
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", ask(l, r));
  }
}
