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 N 500010
#define lowbit(x) x&-x
using namespace std;
int a[N],pos[N],ans[N],fwt[N];
vector<pair<int,int> > v[N];
void update(int x,int p)
{
	for(int i=x;i<N;i+=lowbit(i)) fwt[i]+=p;
}
int sum(int x)
{
	int ss=0;
	for(int i=x;i>0;i-=lowbit(i)) ss+=fwt[i];
	return ss;
}
int main()
{
	int n,q;
	cin>>n>>q;
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i=1;i<=q;i++)
	{
		int l,r;
		cin>>l>>r;
		v[r].push_back({l,i});
	}
	memset(pos,-1,sizeof(pos));
	for(int i=1;i<=n;i++)
	{
		if(pos[a[i]]==-1)
		{
			update(i,1);
			pos[a[i]]=i;
		}
		else
		{
			update(pos[a[i]],-1);
			update(i,1);
			pos[a[i]]=i;
		}
		for(int j=0;j<v[i].size();j++) ans[v[i][j].second]=sum(i)-sum(v[i][j].first-1);
	}
	for(int i=1;i<=q;i++) cout<<ans[i]<<endl;
	return 0;
}