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: Accepted)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
using namespace std;
const int MAXN=5e5+5;
const int M=MAXN*100;
int n,q,tot;
int a[MAXN];
int T[MAXN],lson[M],rson[M],c[M];

int build(int l, int r){
    int root = tot++;
    c[root] = 0;
    if(l != r){
        int mid = (l+r)>>1;
        lson[root] = build(l,mid);
        rson[root] = build(mid+1,r);
    }
    return root;
}

int update(int root,int pos,int val){
    int newroot=tot++,tmp=newroot;
    c[newroot]=c[root]+val;
    int l=1,r=n;
    while(l<r){
        int mid=(l+r)>>1;
        if(pos<=mid){
            lson[newroot]=tot++;
            rson[newroot]=rson[root];
            newroot=lson[newroot];
            root=lson[root];
            r=mid;
        }else{
            rson[newroot]=tot++;
            lson[newroot]=lson[root];
            newroot=rson[newroot];
            root=rson[root];
            l=mid+1;
        }
        c[newroot]=c[root]+val;
    }
    return tmp;
}

int query(int root,int pos){
    int ret=0;
    int l=1,r=n;
    while(pos<r){
        int mid=(l+r)>>1;
        if(pos<=mid){
            r=mid;
            root=lson[root];
        }else{
            ret+=c[lson[root]];
            root=rson[root];
            l=mid+1;
        }
    }
    return ret+c[root];
}

int main(){
    tot = 0;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    T[n+1] = build(1, n);
    map<int,int> mp;
    for(int i = n; i >= 1; i--){
        if(mp.find(a[i])==mp.end()){
            T[i] = update(T[i+1], i, 1);
        }else{
            int tmp = update(T[i+1], mp[a[i]], -1);
            T[i] = update(tmp, i, 1);
        }
        mp[a[i]] = i;
    }
    while(q--){
        int l, r; scanf("%d%d", &l, &r);
        printf("%d\n", query(T[l], r));
    }
    return 0;
}