Jeronimo's List（模拟+计数排序||桶排）

　Jeronimo the bear loves numbers and he is planning to write n numbers in his notebook.

　After writing the first m numbers, Jeronimo felt that he was spending a lot of time thinking new numbers, so he wrote the next n - m missing numbers as the sum modulo 3 × 10⁷ of the numbers in the i - m and i - m + 1 positions for m < i ≤ n

　While Jeronimo was writing, his sister Lupe arrived and asked him q questions. The i - thquestion consist of a number b_i, Jeronimo has to say what would be the number in the position b_iif all the numbers were sorted in ascending order. Jeronimo wants to answer each question as soon as possible but he spends a lot of time counting so he ask your help.

Input

　The first line of the input has three integers n(3 ≤ n ≤ 3 × 10⁷), m (3 ≤ m ≤ min(100, n)) and q(1 ≤ q ≤ 10000).

　The second line contains m numbers a₁, a₂, ..., a_m, (0 ≤ a_i < 3 × 10⁷), The first m numbers that Jeronimo wrote.

　The third line contains q questions b₁, b₂, ..., b_q(1 ≤ b_i ≤ n)

Output

　Print q lines. The i - th line must be the answer of the i - th question made by Lupe.

Examples

Input

6 3 6
1 2 3
1 2 3 4 5 6

Output

1
2
3
3
5
6

Input

10 4 3
1 2 9 10
1 5 10

Output

1
10
30

开始没注意3e7开始对整个数组sort排了下序，T了，

后来二分查第x小的数，T了，又另加了vis数组记忆化标记计算过的，也T了，

愣是没想过桶排

 

计数存，俗称桶排，代码超友好：

 

#include <stdio.h>
#include <string.h>

#define mod 30000000
#define MAX 30000007

int a[MAX], num[MAX];
int re[MAX];

int main()
{
    int n, i, j, pos, m, q, cur;
    scanf("%d %d %d", &n, &m, &q);
    memset(num, 0, sizeof(num));
    
    //数组里的数计数到num数组里
    for(i=0; i<m; i++)
    {
        scanf("%d", &a[i]);
        num[a[i]]++;;
    }
    for(; i<n; i++)
    {
        a[i] = (a[i-m] + a[i-m+1])%mod;
        num[a[i]]++;
    }
    
    //数组re是利用计数数组num的到的排序的后数组
    cur = 0;
    for(i=0;i<MAX;i++)
    {
        for(j=0;j<num[i];j++)
        {
            re[cur++] = i;
        }
    }
    
    while(q--)
    {
        scanf("%d", &pos);
        printf("%d
", re[pos-1]);
    }
    return 0;
}