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 ll long long
#define f first
#define s second
using namespace std;
const int N=1000009;
int bit[N],l[N];
void update(int x,int d){
	while(x<N){
		bit[x]+=d;
		x+=(x & -x);
	}
}
int read(int x){
	int sum=0;
	while(x>0){
		sum+=bit[x];
		x-=(x & -x);
	}
	return sum;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n,q,j=0;
	cin >> n >> q;
	int a[n+1];
	for(int i=1;i<=n;i++)
		cin >> a[i];
	int ans[q];
	pair <pair <int,int> ,int > v[q];
	for(int i=0;i<q;i++){
		cin >> v[i].f.s >> v[i].f.f;
		v[i].s=i;
	}
	sort(v,v+q);
	memset(l,-1,sizeof(l));
	memset(bit,0,sizeof(bit));
	for(int i=1;i<=n;i++){
		if(l[a[i]]!=-1)
			update(l[a[i]],-1);
		l[a[i]]=i;
		update(l[a[i]],1);
		while(v[j].f.f==i){
			ans[v[j].s]=read(v[j].f.f)-read(v[j].f.s-1);
			j++;
		}
	}
	for(int i=0;i<q;i++)
		cout << ans[i] << endl;
}