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 n, Q;
int a[500500];
int bit[500500];
int pre[500500];
void update(int pos, int val){
	while(pos <= n){
		bit[pos] += val;
		pos += (pos & -pos);
	}
}
int query(int pos){
	int sum = 0;
	while(pos > 0){
		sum += bit[pos];
		pos -= (pos & -pos);
	}
	return sum;
}
pair<int, int> qs[500500];
vector<int> q[500500];
int ans[500500];

int main(){
	cin >> n >> Q; 
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 0; i < Q; i++){
		cin >> qs[i].first >> qs[i].second;
		q[qs[i].first].push_back(i);
	}
	memset(pre, -1, sizeof(pre));
	for(int i = n; i >= 1; i--){
		update(i, 1);
		if(pre[a[i]] >= 0) update(pre[a[i]], -1);
		pre[a[i]] = i;
		for(int j = 0; j < q[i].size(); j++){
			ans[q[i][j]] = query(qs[q[i][j]].second);
		}
	}
	for(int i = 0; i < Q; i++) cout << ans[i] << endl;
	return 0;
} 