The Smart Beaver from ABBYY invented a new message encryption method and now wants to check its performance. Checking it manually is long and tiresome, so he decided to ask the ABBYY Cup contestants for help.

A message is a sequence of n integers a1, a2, ..., an. Encryption uses a key which is a sequence of m integers b1, b2, ..., bm (m ≤ n). All numbers from the message and from the key belong to the interval from 0 to c - 1, inclusive, and all the calculations are performed modulo c.

Encryption is performed in n - m + 1 steps. On the first step we add to each number a1, a2, ..., am a corresponding number b1, b2, ..., bm. On the second step we add to each number a2, a3, ..., am + 1 (changed on the previous step) a corresponding number b1, b2, ..., bm. And so on: on step number i we add to each number ai, ai + 1, ..., ai + m - 1 a corresponding number b1, b2, ..., bm. The result of the encryption is the sequence a1, a2, ..., an after n - m + 1 steps.

Help the Beaver to write a program that will encrypt messages in the described manner.

Input

The first input line contains three integers n, m and c, separated by single spaces. 

The second input line contains n integers ai (0 ≤ ai < c), separated by single spaces — the original message. 

The third input line contains m integers bi (0 ≤ bi < c), separated by single spaces — the encryption key.

The input limitations for getting 30 points are: 

  * 1 ≤ m ≤ n ≤ 103
  * 1 ≤ c ≤ 103



The input limitations for getting 100 points are: 

  * 1 ≤ m ≤ n ≤ 105
  * 1 ≤ c ≤ 103

Output

Print n space-separated integers — the result of encrypting the original message.

Examples

Input

4 3 2
1 1 1 1
1 1 1


Output

0 1 1 0


Input

3 1 5
1 2 3
4


Output

0 1 2

Note

In the first sample the encryption is performed in two steps: after the first step a = (0, 0, 0, 1) (remember that the calculations are performed modulo 2), after the second step a = (0, 1, 1, 0), and that is the answer. 
import java.util.Scanner;

public class D177 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int M = in.nextInt();
        int mod = in.nextInt();
        int[] A = new int[N];
        for (int n=0; n<N; n++) {
            A[n] = in.nextInt();
        }
        int[] B = new int[M];
        for (int m=0; m<M; m++) {
            B[m] = in.nextInt();
        }
        for (int m=1; m<M; m++) {
            B[m] += B[m-1];
            B[m] %= mod;
        }
        StringBuilder sb = new StringBuilder();
        for (int n=0; n<N; n++) {
            int toEnd = N-n;
            int missingAtTheEnd = Math.max(0, M-toEnd);
            int missingAtTheStart = Math.max(0, M-1-n);
            int result = B[M-1-missingAtTheStart] + mod - ((missingAtTheEnd == 0) ? 0 :  B[missingAtTheEnd-1]);
            result += A[n];
            result %= mod;
            sb.append(result).append(' ');
        }
        System.out.println(sb);
    }

}
