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 a[maxn], sum[maxn], ans[maxn];
class Poi {
 public:
  int x, y, z;
} query[maxn];
int N, M, gap;
void read() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= M; ++i) {
    scanf("%d%d", &query[i].x, &query[i].y);
    query[i].z = i;
  }
}
inline bool cmp(const Poi& i, const Poi& j) {
  if (i.x / gap != j.x / gap) return i.x / gap < j.x / gap;
  return i.y < j.y;
}
void in(int i, int& tot) {
  if (a[i] > N) return;
  if (sum[a[i]] == a[i]) --tot;
  if (++sum[a[i]] == a[i]) ++tot;
}
void out(int i, int& tot) {
  if (a[i] > N) return;
  if (sum[a[i]] == a[i]) --tot;
  if (--sum[a[i]] == a[i]) ++tot;
}
void deal(int l, int r) {
  memset(sum, 0, sizeof sum);
  int tot = 0;
  for (int i = query[l].x; i <= query[l].y; ++i) in(i, tot);
  ans[query[l].z] = tot;
  for (int i = l + 1; i <= r; ++i) {
    if (query[i].x >= query[i - 1].x)
      for (int j = query[i - 1].x; j < query[i].x; ++j) out(j, tot);
    else
      for (int j = query[i].x; j < query[i - 1].x; ++j) in(j, tot);
    for (int j = query[i - 1].y + 1; j <= query[i].y; ++j) in(j, tot);
    ans[query[i].z] = tot;
  }
}
void getans() {
  gap = (int)sqrt(N);
  sort(query + 1, query + 1 + M, cmp);
  int l = 1;
  for (int i = 2; i <= M; ++i)
    if ((query[i].x / gap != query[i - 1].x / gap)) deal(l, i - 1), l = i;
  deal(l, M);
  for (int i = 1; i <= M; ++i) printf("%d\n", ans[i]);
}
int main() {
  read();
  getans();
}
