coderforce --B. Sereja and Suffixes

B. Sereja and Suffixes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 outm integers l₁, l₂, ..., l_m(1 ≤ l_i ≤ n). For each numberl_i he wants to know how many distinct numbers are staying on the positionsl_i,l_i + 1, ...,n. Formally, he want to find the number of distinct numbers amonga_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 eachl_i.

Input

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

Next m lines contain integers l₁, l₂, ..., l_m. Thei-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

10 101 2 3 4 1 2 3 4 100000 9999912345678910

Output

6666654321

题意：找出输入数组中l后面位置不同的数的个数，满水的一道，可是我竟然没有1A，就是太差劲了。。用一个记录数组arr来记录该数字之前有没有出现过，然后在result数组里面保存不同数字的个数，然后直接查询也就可以了。。我是从后面开始记录的，从前面也可以，不过那就要用最后减去前面那个。。都差不多。。贴下代码：

#include<stdio.h>int arr[100005],sav[100005];int res[100005];int main(){    int n,m;int temp;    for(int i=0;i<100005;i++)    {arr[i]=0;     res[i]=0;     sav[i]=0;    }    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)    {        scanf("%d",&temp);        sav[i]=temp;    }    res[n-1]=1;    arr[sav[n-1]]=1;    for(int i=n-2;i>=0;i--)    {       if(arr[sav[i]]==0)       {           res[i]=res[i+1]+1;           arr[sav[i]]=1;       }       else       res[i]=res[i+1];    }   for(int i=0;i<m;i++)   {       scanf("%d",&temp);       printf("%d\n",res[temp-1]);   }    return 0;}