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.*;
import java.io.*;

public class D1_Encrypting_Messages {

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static FastReader scan = new FastReader();

    public static void main(String[] args) {
        int t = 1;
        while (t-- > 0) {
            solve();
        }
    }

    public static void solve() {
        int n = scan.nextInt();
        int m = scan.nextInt();
        int c = scan.nextInt();

        int[] aArray = new int[n + 1];
        int[] bArray = new int[m + 1];
        for (int i = 1; i <= n; i++) {
            aArray[i] = scan.nextInt();
        }

        for (int i = 1; i <= m; i++) {
            bArray[i] = scan.nextInt();
        }

        for (int i = 1; i <= n - m + 1; i++) {
            int k = i;
            for (int j = 1; j <= m; j++) {
                aArray[k] = (aArray[k] + bArray[j]) % c;
                k++;
            }
        }
        for (int l = 1; l <= n; l++) {
            System.out.print(aArray[l] + " ");
        }
    }
}