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;
#define N 500010
int n,q,a[N],cnt[N],bel[N],l,r,ans[N];
const int B=sqrt(N);
struct rua
{
	int l,r,id;
	void read(int i){scanf("%d%d",&l,&r),id=i;}
	bool operator <(const rua &t)const{return (bel[l]^bel[t.l])?bel[l]<bel[t.l]:((bel[l]&1)?r<t.r:r>t.r);}
}f[N];
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++)
	  f[i].read(i);
	for(int i=1;i<=q;i++)
	  bel[i]=i/B;
	sort(f+1,f+q+1);
	int l=1,r=0,cur=0;
	for(int i=1;i<=q;i++)
	  {
	  int ql=f[i].l,qr=f[i].r;
	  while(l<ql)cur-=!--cnt[a[l++]];
	  while(l>ql)cur+=!cnt[a[--l]]++;
	  while(r<qr)cur+=!cnt[a[++r]]++;
	  while(r>qr)cur-=!--cnt[a[r--]];
	  ans[f[i].id]=cur;
	  }
	for(int i=1;i<=q;i++)
	  printf("%d\n",ans[i]);
	return 0;
}