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.*;

/*
 *	
 *	D.java
 *	Written by Andy Huang: 12:34:23 PM Apr 21, 2012
 */
public class D {
	static void solve() {
		int n = in.nInt();
		int m = in.nInt();
		int c = in.nInt();
		int a[] = new int[n];
		int b[] = new int[m];
		for(int i = 0; i < n; i++)
			a[i] = in.nInt();
		for(int i = 0; i < m; i++)
			b[i] = in.nInt();
		for(int i = 0; i < n-m+1; i++){
			for(int j = i, k = 0; k < m; j++, k++){
				a[j] += b[k];
				a[j] %= c;
			}
		}
		for(int i =0; i < n; i++){
			out.append(a[i]);
			out.append(' ');
		}
		
	}

	public static void main(String[] args) {
		in = new Input();
		out = new Output();
		solve();
		out.print();
	}

	static Output out;
	static Input in;

	static void pln(Object o) {
		System.out.println(o);
	}

	static void pf(Object o) {
		System.out.print(o);
	}


}

final class Input {
	private java.io.BufferedReader reader;
	private java.util.StringTokenizer tokenizer;

	public Input() {
		reader = new java.io.BufferedReader(new java.io.InputStreamReader(System.in));
		tokenizer = null;
	}

	public String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new java.util.StringTokenizer(reader.readLine());
			}
			catch (java.io.IOException e) {
				throw new RuntimeException("I/O Error");
			}
		}
		return tokenizer.nextToken();
	}

	public String nLine() {
		try {
			return reader.readLine();
		}
		catch (java.io.IOException e) {
			throw new RuntimeException(e);
		}
	}

	public long nLong() {
		return Long.parseLong(next());
	}

	public int nInt() {
		return Integer.parseInt(next());
	}

	public double nDouble() {
		return Double.parseDouble(next());
	}

}
final class Output {
	public StringBuilder buffer;

	Output() {
		buffer = new StringBuilder();
	}

	Output(int size) {
		buffer = new StringBuilder(size);
	}

	void print() {
		System.out.print(buffer.toString());
	}

	void flush() {
		System.out.flush();
	}

	<T> void append(T obj) {
		buffer.append(obj);
	}

	<T> void appendln(T obj) {
		append(obj);
		append('\n');
	}

	void delete(int index) {
		buffer.deleteCharAt(index);
	}

}