We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8


Output

1
2
2
1
2
2
6
3
3
3
#include<bits/stdc++.h>
using namespace std;
struct _IO{_IO(){ios::sync_with_stdio(0);cin.tie(0);}}_io;
typedef long long ll; typedef long double db;
const int N = 5e5 + 5, M = 1e9 + 7;

typedef pair<int, int> pii;
#define fi first
#define se second

int bit[N];
void add(int p, int v) {
  for (; p < N; p += p&-p) bit[p] += v;
}
int get(int p) {
  int res = 0;
  for (; p; p -= p&-p) res += bit[p];
  return res;
}
vector<pii> q[N];
int ans[N], lst[N], a[N];

int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int l, r;
  for (int i = 1; i <= m; i++) {
    cin >> l >> r;
    q[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    if (lst[a[i]]) add(lst[a[i]], -1);
    add(i, 1);
    lst[a[i]] = i;
    for (auto &p : q[i]) {
      ans[p.se] = get(i) - get(p.fi-1);
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}
