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;
const int N=500005;
int n,q,a[N],l[N],r[N],b[N],c[N],ans[N];
vector<pair<int,int>>x[N];
void add(int x,int v)
{
	for(int i=x;i<=n;i+=i&(-i))
		b[i]+=v;
}
int sum(int x)
{
	int v=0;
	for(int i=x;i;i-=i&(-i))
		v+=b[i];
	return v;
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=q;i++)
	{
		scanf("%d%d",&l[i],&r[i]);
		x[r[i]].emplace_back(l[i],i);
	}
	for(int i=1;i<=n;i++)
	{
		if(c[a[i]])
			add(c[a[i]],-1);
		c[a[i]]=i;
		add(i,1);
		for(auto [u,v]:x[i])
			ans[v]=sum(i)-sum(u-1);
	}
	for(int i=1;i<=q;i++) printf("%d\n",ans[i]);
	return 0;
}