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 maxn=5e5+5;
const int inf=0x3f3f3f3f;
int c[maxn],vis[maxn],ans[maxn],tree[maxn];
vector<pair<int,int> >v[maxn];
int n,q;
int lowbit(int x)
{
	return x&(-x);
}
void add(int x,int k)
{
	while(x<=n)
	{
		tree[x]+=k;
		x+=lowbit(x);
	}
}
int sum(int x)
{
	int ans=0;
	while(x!=0)
	{
		ans+=tree[x];
		x-=lowbit(x);
	}
	return ans;
}
int main()
{
	cin>>n>>q;
	for(int i=1;i<=n;i++)
		cin>>c[i];
	for(int i=1;i<=q;i++)
	{
		int l,r;
		cin>>l>>r;
		v[r].push_back(make_pair(i,l));
	}
	for(int i=1;i<=n;i++)
	{
		if(vis[c[i]]!=0)
			add(vis[c[i]],-1);
		vis[c[i]]=i;
		add(i,1);
		for(auto j:v[i])
			ans[j.first]=sum(i)-sum(j.second-1);
	}
	for(int i=1;i<=q;i++)
		cout<<ans[i]<<endl;
	return 0;
}
