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=5e4+10;

struct pr {
	int l,r,id,bel;
}q[maxn*20];

int ans[maxn*20],a[maxn*20],n,m,num[1000010],cur;

bool cmp(pr x,pr y) {
	if(x.bel!=y.bel) {
		return x.bel<y.bel;
	}
	return x.r<y.r;
}

void add(int x) {
	num[x] ++;
	if (num[x] == 1)
		cur ++;
}

void del(int x) {
	num[x] --;
	if (num[x] == 0)
		cur --;
}

int main() {
	
	cin >> n;
  cin >> m;
	int blocks = sqrt(n);
	
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= m; ++i) {
		q[i].id = i;
		scanf("%d %d", &q[i].l, &q[i].r);
		q[i].bel = (q[i].l - 1) / blocks  + 1;
	}
	sort(q + 1, q + m + 1, cmp);
	
	
	int l = 1, r = 0;
	
	for (int i = 1; i <= m; ++i) {
		while (l < q[i].l) del(a[l++]);
		while (l > q[i].l) add(a[--l]);
		while (r < q[i].r) add(a[++r]);
		while (r > q[i].r) del(a[r--]);
		ans[q[i].id] = cur;//统计情况；
	}
	
	for (int i =1 ; i <= m; ++i) {
		printf("%d\n",ans[i]);
	}
	return 0;
}