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>
#define rep(i,n) for (long long i = 0; i < (n); ++i)
//g++ -std=c++11 
//
using namespace std;

const int N = 500005;
int n, m;
int c[N];
int maxn;
int sum = 0;
int ans[N];
 
int cnt[N];
 
struct query {
  int l, r, id;
  bool operator<(const query &x) const {
    if (l / maxn != x.l / maxn) return l < x.l;
    return (l / maxn) & 1 ? r < x.r : r > x.r;
  }
} a[N];
 
inline void add(int i){
	if(cnt[i]==0){
		sum +=1;
	}
	cnt[i]++;
}
inline void del(int i){
	if(cnt[i]==1){
		sum -= 1;
	}
	cnt[i]--;
}
 
 
int main (){
	#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
	IOS
	cin >> n >> m;
	for(long long i = 1; i <= n; i++) {
		cin >> c[i];
	}
	maxn = sqrt(n);
	rep(i, m) {
		cin >> a[i].l >> a[i].r;
		a[i].id = i;
	}
	sort(a,a+m);
	for(int i = 0,l=1,r=0; i<m; i++){
		while(r > a[i].r) del(c[r--]);
		while(r < a[i].r) add(c[++r]);
		while (l > a[i].l) add(c[--l]);
		while (l < a[i].l) del(c[l++]);
 
	    ans[a[i].id] = sum;
	}
	for(int i = 0; i < m; i++){
		cout << ans[i] << endl;
	}
}
