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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n,m,t;
struct node
{
	int l,r,no;
	bool operator<(const node s)const
	{
		return r<s.r;
	}
}a[500005];
int d[500005],c[500005],p[2000005],ans[500005];
int lowbit(int x)
{
	return x&(-x);
}
int sum(int x)
{
	int res=0;
	while(x>0)
	{
		res+=d[x];
		x-=lowbit(x);
	}
	return res;
}
void add(int x,int lazy)
{
	while(x<=n)
	{
		d[x]+=lazy;
		x+=lowbit(x);
	}
	return;
}
int main()
{
	scanf("%d%d",&n,&t);
	for(int i=1;i<=n;i++)
		scanf("%d",&c[i]);
	for(int i=1;i<=t;i++)
	{
		a[i].no=i;
	    scanf("%d%d",&a[i].l,&a[i].r);
	}
	sort(a+1,a+t+1);
	int beg=1;
	for(int i=1;i<=t;i++)
	{
		for(int j=beg;j<=a[i].r;j++)
		{
			if(!p[c[j]])
			    add(j,1);
			else
			{
				add(p[c[j]],-1);
				add(j,1);
			}
			p[c[j]]=j;
		}
		beg=a[i].r+1;
		ans[a[i].no]=sum(a[i].r)-sum(a[i].l-1);
	}
	for(int i=1;i<=t;i++)
	    printf("%d\n",ans[i]);
	return 0;
}