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 double eps = 1e-8;
int n, q;
int a[120007];
pair<pair<int, int>, int> b[120007];
map<int, int> has;
bool cmp(const pair<pair<int, int>, int> &a,
         const pair<pair<int, int>, int> &b) {
  return a.first.second < b.first.second;
}
int BIT[120007];
vector<int> vec[120007];
int ans[120007];
void insert(int x, int val) {
  ++x;
  while (x <= n + 2) {
    BIT[x] += val;
    x += x & -x;
  }
}
int sum(int x) {
  ++x;
  int ret = 0;
  while (x) {
    ret += BIT[x];
    x -= x & -x;
  }
  return ret;
}
int main() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) has[a[i]] = 1;
  int cur = 0;
  for (typeof(has.begin()) it = has.begin(); it != has.end(); it++)
    it->second = cur++;
  for (int i = 0; i < q; i++)
    scanf("%d%d", &b[i].first.first, &b[i].first.second);
  for (int i = 0; i < q; i++) b[i].second = i;
  sort(b, b + q, cmp);
  int pos = 0;
  for (int i = 0; i < n; i++) {
    int idx = has[a[i]];
    if (vec[idx].size() >= a[i] + 1)
      insert(vec[idx][vec[idx].size() - a[i] - 1], 1);
    if (vec[idx].size() >= a[i]) insert(vec[idx][vec[idx].size() - a[i]], -1);
    vec[idx].push_back(i);
    if (vec[idx].size() >= a[i]) insert(vec[idx][vec[idx].size() - a[i]], 1);
    if (vec[idx].size() >= a[i] + 1)
      insert(vec[idx][vec[idx].size() - a[i] - 1], -1);
    while (pos < q && b[pos].first.second == i + 1) {
      ans[b[pos].second] =
          sum(b[pos].first.second - 1) - sum(b[pos].first.first - 2);
      ++pos;
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
  return 0;
}
