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: Runtime Error)

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 5000010
#define maxm 200010
struct node
{
    int l,r,no;
}c[maxm];
int n,m;
int tree[maxn],front[maxn],last[maxn];
int ans[maxm];
bool cmp(node x,node y)
{
    return x.r<y.r;
}
int lowbit(int i)
{
    return i&(-i);
}
void update(int i,int x)
{
    while(i<=maxn)
    {
        tree[i]+=x;
        i+=lowbit(i);
    }
}
int query(int i)
{
    int sum=0;
    while(i>0)
    {
        sum+=tree[i];
        i-=lowbit(i);
    }
    return sum;
}
int main()
{   
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        int temp;
        scanf("%d",&temp);
        front[i]=last[temp]+1;//front数组为i位置上的数上一次出现的位置+1
        last[temp]=i;//last数组为i位置上的数到目前为止最后一次出现的位置
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&c[i].l,&c[i].r);
        c[i].no=i;//每次查询的序号
    }
    sort(c+1,c+1+m,cmp);//按右端点排序
    int k=1;//k初值为1
    for(int i=1;i<=m;i++)
    {
        while(k<=c[i].r)
        {
            k++;
            update(front[k-1],1);
            update(k,-1);
        }
        ans[c[i].no]=query(c[i].l);//查询左端点
    }
    for(int i=1;i<=m;i++)
    printf("%d\n",ans[i]);
    return 0;
}