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 = 500005;

int n, q, c[MAXN], li, ri, res[MAXN];
vector < pair <int, int> > qur[MAXN];
vector <int> v[MAXN], add[MAXN];

int tree[MAXN];
void tadd(int u, int val){
	u++;
	while(u < MAXN){
		tree[u] += val;
		u += u & (-u);
	}
}
int sum(int u){
	u++;
	int ret = 0;
	while(u > 0){
		ret += tree[u];
		u -= u & (-u);
	}
	return ret;
}

int main(){
	scanf("%d %d", &n, &q);
	for(int i = 0; i < n; i++){
		scanf("%d", &c[i]);
		v[c[i]].push_back(i);
	}
	
	for(int i = 1; i <= n; i++){
		for(int j = 0; j + 1 < (int)v[i].size(); j++){
			add[v[i][j + 1]].push_back(v[i][j]);
		}
	}
	
	for(int i = 0; i < q; i++){
		scanf("%d %d", &li, &ri);
		li--, ri--;
		
		qur[ri].push_back(make_pair(li, i));
	}
	
	for(int i = 0; i < n; i++){
		for(int j : add[i]){
			tadd(j, 1);
		}
		for(pair <int, int> qi : qur[i]){
			res[qi.second] = i - qi.first + 1 - (sum(i) - sum(qi.first - 1));
		}
	}
	
	for(int i = 0; i < q; i++)
		printf("%d\n", res[i]);
	
	return 0;
}
