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>
using namespace std;
#define maxn 500010

int n,m,a[maxn];
struct que{int x,y,pos;}ask[maxn];
int last[maxn],ans[maxn];
int tr[maxn];
void add(int x,int y){for(;x<=maxn-5;x+=(x&-x))tr[x]+=y;}
void Add(int x,int y){add(x,1);add(y+1,-1);};
int sum(int x){int re=0;for(;x>=1;x-=(x&-x))re+=tr[x];return re;}
bool cmp(que x,que y){return x.y<y.y;}

int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=m;i++)scanf("%d %d",&ask[i].x,&ask[i].y),ask[i].pos=i;
	sort(ask+1,ask+m+1,cmp);
	int now=1;for(int i=1;i<=n;i++){
		Add(last[a[i]]+1,i);last[a[i]]=i;
		while(now<=m&&ask[now].y==i)ans[ask[now].pos]=sum(ask[now].x),now++;
		if(now>m)break;
	}
	for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
}