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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class D {
	BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = null;

	private void solution() throws IOException {
		int n = nextInt();
        int m=nextInt();
        int c=nextInt();
        int a[]=new int[n];
        int b[]=new int[m];
        int t=0;
        for(int i=0;i<n;i++){a[i]=nextInt();}
        for(int i=0;i<m;i++){b[i]=nextInt();}
    for(int i=0;i<n-m+1;i++){
    	for(int j=t;j<m+t;j++){
    		a[j]+=b[j-t];
    		a[j]%=c;
    	}
    	t++;
    }
    for(int i=0;i<n;i++){System.out.print(a[i]+" ");}
		}

	
	
	

	String nextToken() throws IOException {
		if (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(bf.readLine());
		}
		return st.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	public void print(int a[]) {
		for (int i = 0; i < a.length; i++) {
			System.out.print(a[i] + " ");
		}
	}

	public static void main(String args[]) throws IOException {
		new D().solution();
	}
}