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<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
#define inf 1000000000
#define mod 1000000007
#define eps 1e-10
#define N 500005
#define rep(i, l, r) for(int i=l; i<r; ++i)
const double pi=acos(-1.0);

using namespace std;
int n, q;
int c[N];
struct query{
	int num, l, r;
	bool operator < (const query &rhs) const{
		return l/700==rhs.l/700 ? r<rhs.r : l<rhs.l;
	}
}a[N];
int cnt[N];
int ans[N];
int main(){
	scanf("%d%d", &n, &q);
	rep(i, 1, n+1) scanf("%d", &c[i]);
	rep(i, 1, q+1){
		a[i].num=i;
		scanf("%d%d", &a[i].l, &a[i].r);
	}
	sort(a+1, a+q+1);
	int l=1, r=0, tmp=0;
	rep(i, 1, q+1){
		while(l>a[i].l) if(cnt[c[--l]]++==0) ++tmp;
		while(r<a[i].r) if(cnt[c[++r]]++==0) ++tmp;
		while(l<a[i].l) if(--cnt[c[l++]]==0) --tmp;
		while(r>a[i].r) if(--cnt[c[r--]]==0) --tmp;
		ans[a[i].num]=tmp;
	}
	rep(i, 1, q+1) printf("%d\n", ans[i]);
	return 0;
}