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 <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#define LL long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
const int maxn=1145141;
int a[maxn],tree[maxn],h[maxn],ans[maxn];
int N,q;
struct Node{
	int l,r,pos;
};
Node que[maxn];
bool cmp(Node x,Node y){
	return x.r<y.r;
}
int lowbit(int n){
	return n&(-n);
}
void add(int n,int now){
	while(n<=N){
		tree[n]+=now;
		n+=lowbit(n);
	}
}
int sum(int n){
	int ret=0;
	while(n!=0){
		ret+=tree[n];
		n-=lowbit(n);
	}
	return ret;
}
int nxt=1;
int main(){
	cin>>N>>q;
	for(int i=1;i<=N;i++)cin>>a[i];
	for(int i=1;i<=q;i++){
		cin>>que[i].l>>que[i].r;
		que[i].pos=i;
	}
	sort(que+1,que+1+q,cmp);
	for(int i=1;i<=q;i++){
		for(int j=nxt;j<=que[i].r;j++){
			if(h[a[j]])add(h[a[j]],-1);
			add(j,1);
			h[a[j]]=j;
		}
		nxt=que[i].r+1;
		ans[que[i].pos]=sum(que[i].r)-sum(que[i].l-1);
	}
	for(int i=1;i<=q;i++)cout<<ans[i]<<endl;
	return 0;
}
