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
//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5+5;

int a[N], b[N], book[N], ans[N], n, q;

struct Node {
    int l, r, id;
    bool operator < (const Node &a) const {
		return r < a.r;
	}
}p[N];
void add(int n, int u) {
	for(;n<=N;n+=n&(-n)) {
		b[n] += u;
	}
}
int sum(int n) {
	int res = 0;
	for(;n;n-=n&(-n)) {
		res += b[n];
	}
	return res;
}

int main()
{
	scanf("%d%d", &n, &q);
	for(int i=1;i<=n;i++) {
		scanf("%d", &a[i]);
	}
	for(int i=1;i<=q;i++) {
		scanf("%d%d", &p[i].l, &p[i].r);
		p[i].id = i;
	}
	sort(p+1, p+1+q);
//	puts("Sorted. ");
	int nxt = 1;
    for(int i=1;i<=q;i++) {
//   	printf("For: i=%d\n", i);
		for(int j=nxt;j<=p[i].r;j++) {
//			printf("\tj=%d\n", j);
			if(book[a[j]]) {
				add(book[a[j]], -1);
			}
			add(j, 1);
			book[a[j]] = j;
		}
		nxt = p[i].r + 1;
		ans[p[i].id] = sum(p[i].r) - sum(p[i].l-1);
	}
	for(int i=1;i<=q;i++) {
		printf("%d\n", ans[i]);
	}
	return 0;
}