Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4 3
1 2 1 3
1 3
2 4
3 3

Sample Output 1

2
3
1

The 1-st, 2-nd, and 3-rd balls from the left have the colors 1, 2, and 1 - two different colors.

The 2-st, 3-rd, and 4-th balls from the left have the colors 2, 1, and 3 - three different colors.

The 3-rd ball from the left has the color 1 - just one color.

Sample Input 2

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

Sample Output 2

1
2
2
1
2
2
6
3
3
3

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10;
int bl,t,n,m;
int res,a[N],cnt[N],ans[N];
int curR,curL;
struct node
{
  int l,r,p;
  bool operator <(const node &W)const
  {
    return l<W.l;
  }
}e[N];
bool cmp(const node &a,const node &b)
{
  return (a.l/bl)==(b.l/bl) ?  a.r<b.r : a.l>b.l;
}
void remove(int i)
{
  if((--cnt[a[i]])==0) res--;
}
void add(int i)
{
  if((++cnt[a[i]])==1) res++;
}
int main()
{

     memset(ans,0,sizeof(ans));
     memset(cnt,0,sizeof(cnt));
     scanf("%d %d",&n,&m);
     for(int i=1;i<=n;i++) scanf("%d",&a[i]);

     bl=sqrt(n);
     for(int i=1;i<=m;i++)
     {
       scanf("%d %d",&e[i].l,&e[i].r);
       e[i].p=i;
     }
     sort(e+1,e+1+m,cmp);
     res=0;
     curL=1;curR=0;
     for(int i=1;i<=m;i++)
     {
       int L=e[i].l,R=e[i].r;
       while(curL<L) remove(curL++);
       while(curL>L) add(--curL);
       while(curR<R) add(++curR);
       while(curR>R) remove(curR--);
       ans[e[i].p]=res;
     }
     for(int i=1;i<=m;i++) printf("%d\n",ans[i]);

}