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<cstring>
#include<algorithm>

#define lowbit(X) ((X)&(-(X)))
using namespace std;
struct Query
{
	int l,r,id;
	bool operator < (const Query& rhs)const
	{
		return r<rhs.r;
	}
};
const int N=505050;
int n,q,sum[N],last[N * 10],ans[N],a[N];
Query Q[N];
void modify(int pos,int ope)
{
	if(ope) for(int i=pos;i<=n;i+=lowbit(i)) sum[i]+=1;
	else for(int i=pos;i<=n;i+=lowbit(i)) sum[i]-=1;
}
int query(int pos)
{
	int res=0;
	for(int i=pos;i>0;i-=lowbit(i)) res+=sum[i];
	return res;
}
void init() {
	scanf("%d",&n);
	scanf("%d",&q);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=0;i<q;i++)
		scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id=i;
	sort(Q,Q+q);
}
int main()
{
	init();
	for(int i=1,k=0;i<=n;i++)
	{
		if(last[a[i]]) modify(last[a[i]],0);
		modify(i,1);
		last[a[i]]=i;
		while(Q[k].r==i)
			ans[Q[k].id]=query(i)-query(Q[k].l-1),k++;
	}
	for(int i=0;i<q;i++)printf("%d\n",ans[i]);
	return 0;
}
