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;
#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("O3")
template <class T>
inline T pw(T a, long long n, T res = 1) {
  while (n) {
    if (n & 1) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}
inline long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = (s << 1) + (s << 3) + ch - '0';
    ch = getchar();
  }
  return s * f;
}
inline long long fib(long long n) {
  double a = pw((1.0 + double(sqrt(5))), n);
  double b = pw((1.0 - double(sqrt(5))), n);
  double c = pw(2.0, n) * (double)sqrt(5);
  return ceil((a - b) / c);
}
const long long N = 2e5 + 20;
const long long E = 1e6 + 20;
const long long inf = 1e18;
const long long md = 1e9 + 7;
bool e = 1;
long long n, m, a[N], answ[N], cod = 450;
struct A {
  long long l, r, id;
} p[N];
bool cmp(A x, A y) {
  if (x.l / cod != y.l / cod) return x.l / cod < y.l / cod;
  return x.r < y.r;
}
bitset<N> ans;
long long cnt[N];
void add(long long x) {
  if (a[x] > n) return;
  cnt[a[x]]++;
  if (cnt[a[x]] == a[x]) {
    ans[a[x]] = 1;
  } else {
    ans[a[x]] = 0;
  }
}
void del(long long x) {
  if (a[x] > n) return;
  cnt[a[x]]--;
  if (cnt[a[x]] == a[x]) {
    ans[a[x]] = 1;
  } else {
    ans[a[x]] = 0;
  }
}
signed main() {
  if (e) ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) cin >> p[i].l >> p[i].r, p[i].id = i;
  sort(p + 1, p + m + 1, cmp);
  long long L = 1, R = 0;
  for (long long i = 1; i <= m; i++) {
    while (L < p[i].l) del(L++);
    while (L > p[i].l) add(--L);
    while (R > p[i].r) del(R--);
    while (R < p[i].r) add(++R);
    answ[p[i].id] = ans.count();
  }
  for (long long i = 1; i <= m; i++) cout << answ[i] << "\n";
}
