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 bi, Jeronimo has to say what would be the number in the position biif 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, ..., am, (0 ≤ ai < 3 × 107), The first m numbers that Jeronimo wrote.
The third line contains q questions b1, b2, ..., bq(1 ≤ bi ≤ n)
Output
Print q lines. The i - th line must be the answer of the i - th question made by Lupe.
Examples
6 3 61 2 31 2 3 4 5 6
123356
10 4 31 2 9 101 5 10
11030
1 #include <stdio.h>
2 #include <string.h>
3
4 #define mod 30000000
5 #define MAX 30000007
6
7 int a[MAX], num[MAX];
8 int re[MAX];
9
10 int main()
11 {
12 int n, i, j, pos, m, q, cur;
13 scanf("%d %d %d", &n, &m, &q);
14 memset(num, 0, sizeof(num));
15
16 //数组里的数计数到num数组里
17 for(i=0; i<m; i++)
18 {
19 scanf("%d", &a[i]);
20 num[a[i]]++;;
21 }
22 for(; i<n; i++)
23 {
24 a[i] = (a[i-m] + a[i-m+1])%mod;
25 num[a[i]]++;
26 }
27
28 //数组re是利用计数数组num的到的排序的后数组
29 cur = 0;
30 for(i=0;i<MAX;i++)
31 {
32 for(j=0;j<num[i];j++)
33 {
34 re[cur++] = i;
35 }
36 }
37
38 while(q--)
39 {
40 scanf("%d", &pos);
41 printf("%d\n", re[pos-1]);
42 }
43 return 0;
44 }