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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int N,Q;
const int W=700;
int c[5<<17];
int ans[5<<17];
int cnt[5<<17];
int main()
{
	scanf("%d%d",&N,&Q);
	for(int i=0;i<N;i++)scanf("%d",&c[i]);
	vector<vector<pair<int,pair<int,int> > > >qs((N+W-1)/W);
	for(int i=0;i<Q;i++)
	{
		int l,r;scanf("%d%d",&l,&r);
		l--;
		qs[l/W].push_back(make_pair(r,make_pair(l,i)));
	}
	int L=0,R=0;
	int all=0;
	for(int i=0;i<qs.size();i++)
	{
		while(L<i*W&&L<N)
		{
			if(!--cnt[c[L++]])all--;
		}
		while(i*W<R)
		{
			if(!--cnt[c[--R]])all--;
		}
		sort(qs[i].begin(),qs[i].end());
		for(int j=0;j<qs[i].size();j++)
		{
			int r=qs[i][j].first;
			while(R<r)
			{
				if(!cnt[c[R++]]++)all++;
			}
			int l=qs[i][j].second.first;
			while(L<l)
			{
				if(!--cnt[c[L++]])all--;
			}
			while(L>l)
			{
				if(!cnt[c[--L]]++)all++;
			}
			ans[qs[i][j].second.second]=all;
		}
	}
	for(int i=0;i<Q;i++)printf("%d\n",ans[i]);
}
