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 N=5e+5+5;
vector<int> st[4*N];
int a[N],c[N],b[N],n;

void build(int l=1,int r=n,int v=1)
{
	if(l==r)
	{
		st[v].push_back(a[l]);
		return;
	}
	build(l,(l+r)/2,2*v);
	build((l+r)/2+1,r,2*v+1);
	merge(st[2*v].begin(),st[2*v].end(),st[2*v+1].begin(),st[2*v+1].end(),back_inserter(st[v]));
}

int querry(int l,int r,int x,int tl=1,int tr=n,int v=1)
{
	if(l>r)
		return 0;
	if(l==tl&&r==tr)
	{
		int ans=upper_bound(st[v].begin(),st[v].end(),x)-st[v].begin();
		int y=st[v].size();
		ans=y-ans;
		return ans;

	}
	int tm=(tl+tr)/2;
	auto lft=querry(l,min(tm,r),x,tl,tm,2*v);
	auto rgt=querry(max(l,tm+1),r,x,tm+1,tr,2*v+1);
	int ans=lft+rgt;
	return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	int q;
	cin>>n>>q;
	for(int i=1;i<=n;i++)
	{
		b[i]=INT_MAX;
		cin>>c[i];
	}
	for(int i=n;i>=1;i--)
	{
		a[i]=b[c[i]];
		b[c[i]]=i;
	}
	build();
	while(q--)
	{
		int l,r;
		cin>>l>>r;
		cout<<querry(l,r,r)<<"\n";	
	}
	
}



