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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

const int B=800;

struct query{
	int l,r;
	int id;
	query(){}
	query(int l1,int r1,int ii){
		l=l1;
		r=r1;
		id=ii;
	}
	bool operator<(const query &q)const{
		if(l/B==q.l/B){
			return r<q.r;
		}
		return l/B<q.l/B;
	}
};

int n,q;
int c[500005];
int cnt[500005];
int cur=0;
int ans[500005];
query qq[500005];

void add(int pos){
	cnt[c[pos]]++;
	if(cnt[c[pos]]==1)cur++;
}

void dec(int pos){
	cnt[c[pos]]--;
	if(cnt[c[pos]]==0)cur--;
}

int main(void){
	scanf("%d%d",&n,&q);
	for(int i=0;i<n;i++){
		scanf("%d",&c[i]);
	}
	for(int i=0;i<q;i++){
		scanf("%d%d",&qq[i].l,&qq[i].r);
		qq[i].l--;
		qq[i].id=i;
	}
	sort(qq,qq+q);
	int l=0,r=0;
	for(int i=0;i<q;i++){
		while(qq[i].l<l){
			l--;
			add(l);
		}
		while(qq[i].r>r){
			add(r);
			r++;
		}
		while(qq[i].l>l){
			dec(l);
			l++;
		}
		while(qq[i].r<r){
			r--;
			dec(r);
		}
		ans[qq[i].id]=cur;
	}
	for(int i=0;i<q;i++){
		printf("%d\n",ans[i]);
	}
	return 0;
}
