CF--368B

Sereja and Suffixes

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

Input

10 101 2 3 4 1 2 3 4 100000 9999912345678910

Output

6666654321

解体思路:题意为查询ali到an有多少个不同的元素，因为区间的结尾都是an所以我们可以反过来，从结尾开始统计到ali一共有多少个不同的元素，遍历一遍就可以了，所以这样就不会超时了。

代码如下：

#include<stdio.h>#include<string.h>int vist[100010];int a[100010];int ans[100010];int main(){int n,m,i,j,k,sum;while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++) scanf("%d",&a[i]);sum=0;for(i=n;i>0;i--){if(!vist[a[i]]){sum++;vist[a[i]]=1;}ans[i]=sum;}for(i=1;i<=m;i++){scanf("%d",&k);printf("%d\n",ans[k]);}}return 0;}