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;
constexpr int Bsize=1000;


int N,Q,C[500000],L[500000],R[500000],res[500000],b[500000],c[500000],tree[1<<20];
vector<int>q[500000];

void add(int x){
	x+=(1<<19);
	while(x){
		tree[x]++;
		x/=2;
	}
}
void erase(int x){
	if(x==-1)return;
	x+=(1<<19);
	while(x){
		tree[x]--;
		x/=2;
	}
}

int calc(int l,int r){
	int s=0;
	l+=(1<<19);
	r+=(1<<19);
	while(l<r){
		if(l&1)s+=tree[l];
		if(~r&1)s+=tree[r];
		l=(l+1)/2;
		r=(r-1)/2;
	}
	if(l==r)s+=tree[l];
	return s;
}

int main(){
	cin>>N>>Q;
	for(int i=0;i<N;i++){
		cin>>C[i];
		C[i]--;
	}
	for(int i=0;i<Q;i++){
		cin>>L[i]>>R[i];
		L[i]--;
		R[i]--;
		q[R[i]].push_back(i);
	}
	for(int i=0;i<N;i++){
		c[i]=-1;
	}
	for(int i=0;i<N;i++){
		b[i]=c[C[i]];
		c[C[i]]=i;
	}
	for(int i=0;i<N;i++){
		add(i);
		erase(b[i]);
		for(int j:q[i]){
			res[j]=calc(L[j],R[j]);
		}
	}
	for(int i=0;i<Q;i++){
		cout<<res[i]<<endl;
	}
}
