Codeforce Round #215 Div2 B

B. Sereja and Suffixes

time limit per test 1 second

memory limit per test 256 megabytes

 

Sereja has an array a, consisting of n integers a₁, a₂, ..., a_n. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out m integers l₁, l₂, ..., l_m (1 ≤ l_i ≤ n). For each number l_i he wants to know how many distinct numbers are staying on the positions l_i, l_i + 1, ..., n. Formally, he want to find the number of distinct numbers among a_li, a_li + 1, ..., a_n.?

Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each l_i.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 10⁵). The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵) — the array elements.

Next m lines contain integers l₁, l₂, ..., l_m. The i-th line contains integer l_i (1 ≤ l_i ≤ n).

Output

Print m lines — on the i-th line print the answer to the number l_i.

Sample test(s)

Input

10 10 
1 2 3 4 1 2 3 4 100000 99999 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10

Output

6 
6 
6 
6 
6 
5 
4 
3 
2 
1

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <queue>

using namespace std;
#define INF 0x7fffffff
#define maxm 1001
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
//#define y1 y8687969
//#define fill(x,y) memset(x,y,sizeof(x))

typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }

const int maxn = 110005;

int n, m, t, s, k, x;
int a[maxn], b[maxn],vis[maxn];
int main(){
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++)scanf("%d", &a[i]);
    b[n - 1] = 1;
    vis[a[n - 1]] = 1;
    for (int i = n - 2; i >= 0; i--){
        if (vis[a[i]])b[i] = b[i + 1];
        else b[i] = b[i + 1] + 1;
        vis[a[i]] = 1;
    }
    while (k--){
        scanf("%d", &x);
        printf("%d\n", b[x - 1]);
    }
    return 0;
}