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=500009;

int n,m;
int a[maxn];
int nex[maxn],tong[maxn];

int ql[maxn],qr[maxn];
vector<int>G[maxn];
int ans[maxn];

int c[maxn];
int lowbit(int x){
	return x&(-x);
}
void Add(int x,int val){
	while(x<=n){
		c[x]+=val;x+=lowbit(x);
	}
}
int query(int x){
	int ret=0;
	while(x){
		ret+=c[x];x-=lowbit(x);
	}
	return ret;
}

int main(){
	cin>>n>>m;
	for(int i=1;i<=n;++i)scanf("%d",&a[i]);
	for(int i=n;i>=1;--i){
		nex[i]=tong[a[i]];
		tong[a[i]]=i;
	}
	for(int i=1;i<=m;++i){
		scanf("%d%d",&ql[i],&qr[i]);
		G[ql[i]].push_back(i);
	}
	for(int i=1;i<=n;++i)if(tong[i])Add(tong[i],1);
	for(int i=1;i<=n;++i){
		for(int j=0;j<G[i].size();++j){
			int id=G[i][j];
			ans[id]=query(qr[id]);
		}
		Add(i,-1);
		if(nex[i])Add(nex[i],1);
	}
	for(int i=1;i<=m;++i)printf("%d\n",ans[i]);
	return 0;
}