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;
int a[210000], f[210000], ans[210000];
pair<pair<int, int>, int> b[210000];
int n, m;
map<int, deque<int> > g;
void ins(int x, int v) {
  while (x) f[x] += v, x -= x & -x;
}
void ins(int x, int y, int v) {
  ins(x - 1, -v);
  ins(y, v);
}
void insert(int i, int x) {
  deque<int> &a = g[x];
  a.push_back(i);
  if (int((a).size()) < x) {
    return;
  }
  if (int((a).size()) == x) {
    ins(1, a[0], 1);
    return;
  }
  if (int((a).size()) == x + 1) {
    ins(1, a[0], -1);
    ins(a[0] + 1, a[1], 1);
    return;
  }
  if (int((a).size()) == x + 2) {
    ins(a[0] + 1, a[1], -1);
    ins(a[1] + 1, a[2], 1);
    a.pop_front();
    return;
  }
}
int getval(int x) {
  int ret = 0;
  while (x <= n) ret += f[x], x += x & -x;
  return ret;
}
int main() {
  memset(f, 0, sizeof(f));
  g.clear();
  cin >> n >> m;
  for (int i = (1); i <= (n); i++) scanf("%d", a + i);
  for (int i = (1); i <= (m); i++)
    scanf("%d%d", &b[i].first.second, &b[i].first.first), b[i].second = i;
  sort(b + 1, b + m + 1);
  int cur = 0;
  for (int i = (1); i <= (m); i++) {
    for (int j = (cur + 1); j <= (b[i].first.first); j++) insert(j, a[j]);
    cur = b[i].first.first;
    ans[b[i].second] = getval(b[i].first.second);
  }
  for (int i = (1); i <= (m); i++) printf("%d\n", ans[i]);
  return 0;
}
