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
// segment tree/BIT number of unique element in a range
#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int mx=500010;
int arr[mx],n,m,last[mx];
pair<int,int> que[mx];
vector<int> g[mx];
int BIT[mx];
void update(int pos,int val)
{
	for(;pos<=mx;pos+=pos&(-pos))BIT[pos]+=val;
}
int query(int pos)
{
	int sum=0;
	for(;pos>0;pos-=pos&(-pos))sum+=BIT[pos];
	return sum;
}
map<pair<int,int>,int> mp;

int main(){
	fast;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>arr[i];
	}
	for(int i=0;i<m;i++){
		cin>>que[i].x>>que[i].y;
		g[que[i].y].push_back(que[i].x);
	}
	for(int i=1;i<=n;i++){
		if(last[arr[i]]){
			update(last[arr[i]],-1);
		}
		last[arr[i]]=i;
		update(i,1);
		for(auto x:g[i]){
			mp[{x,i}]=(query(i)-query(x-1));
		}
	}
	for(int i=0;i<m;i++){
		cout<<mp[que[i]]<<endl;
	}
}
