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;
int bu;
struct q{
	int l , r , in;
};
bool comp(q a , q b)
{
	if((a.l)/bu!=(b.l)/bu)
		return (a.l)/bu < (b.l)/bu;
	else
		return a.r < b.r;
}
int main()
{
	ios_base :: sync_with_stdio(false);
	cin.tie(0);
	int n , m1;
	cin>>n>>m1;
	int arr[n] , answer[m1];
	for(int i=0;i<n;i++)
		cin>>arr[i];
	q query[m1+1];
	bu=sqrt(n);
	for(int i=0;i<m1;i++)
	{
		int x , y;
		cin>>x>>y;
		x-- , y--;
		query[i]={x , y , i};
	}
	sort(query , query+m1 , comp);
	int m[n+1] , total=0;
	memset(m , 0 , sizeof(m));
	int curl=0 , curr=-1;
	for(int i=0;i<m1;i++)
	{
		while(curr<query[i].r)
		{
			if(m[arr[curr+1]]==0)
				total++;
			m[arr[++curr]]++;
		}
		while(curl<query[i].l)
		{
				m[arr[curl]]--;
				if(m[arr[curl]]==0)
					total--;
				curl++;
		}
		while(curl>query[i].l)
		{
			if(m[arr[curl-1]]==0)
				total++;
			m[arr[--curl]]++;
		}
		while(curr>query[i].r)
		{
			m[arr[curr]]--;
			if(m[arr[curr]]==0)
				total--;
			curr--;
		}
		answer[query[i].in]=total;
	}
	for(int i=0;i<m1;i++)
		cout<<answer[i]<<" ";
}