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<stdio.h>
#include<math.h>
int size;
struct info {
int L;
int R;
int e;
};
int com(struct info i,struct info j)
{
    if((i.L/size)<(j.L/size)) return 1;
    if((i.L/size)>(j.L/size)) return 0;
    if((i.R)<(j.R)) return 1;
    return 0;
}

int main()
{
    int n,i,j,m,cs,cl,cr,k,l;
    scanf("%d %d",&n,&m);
    size=sqrt(n);
    int a[n],c[500001],b[m][3];
    struct info q[m];
    for(i=0;i<=500000;i++) c[i]=0;
    for(i=0;i<n;i++) scanf("%d",&a[i]);
    for(i=0;i<m;i++)
    {
        scanf("%d %d",&b[i][0],&b[i][1]);
        q[i].L=b[i][0]-1;
        q[i].R=b[i][1]-1;
        q[i].e=i;
    }
    for(i=0;i<m-1;i++)
    {
        k=i;
        for(j=i+1;j<m;j++)
        {
            if(com(q[k],q[j])==0) k=j;
        }
        if(k!=i)
        {
            struct info temp;
            temp=q[i];
            q[i]=q[k];
            q[k]=temp;
        }
    }
    cl=0;
    cr=0;
    cs=0;
    for(i=0;i<m;i++)
    {
        while(q[i].L>cl)
        {
            c[a[cl]]--;
            if(c[a[cl]]==0) cs--;
            cl++;
        }
        while(q[i].L<cl)
        {
          c[a[cl-1]]++;
            if(c[a[cl-1]]==1) cs++;
            cl--;
        }
        while(q[i].R>=cr)
        {
            c[a[cr]]++;
            if(c[a[cr]]==1) cs++;
            cr++;
        }
        while(q[i].R+1<cr)
        {
            c[a[cr-1]]--;
            if(c[a[cr-1]]==0) cs--;
            cr--;
        }
        int u=q[i].e;
        b[u][2]=cs;
    }
    for(i=0;i<m;i++) printf("%d\n",b[i][2]);
    return 0;
}