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.*;
import java.util.*;

public class D {

	BufferedReader in;
	StringTokenizer st;
	PrintWriter out;

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

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

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

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

	void solve() throws Exception {
		int n = nextInt(), m = nextInt(), mod = nextInt();
		long a[] = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		long b[] = new long[m];
		for (int i = 0; i < m; i++)
			b[i] = nextLong();

		long curr = 0;
		for (int i = 0; i < n; i++) {
			if (i < m)
				curr += b[i];
			int p = m - n - 1 + i;
			if (p < m && p >= 0)
				curr -= b[p];
			a[i] += curr;
		}
		for (int i = 0; i < n; i++)
			out.print(a[i] % mod + " ");
	}

	void run() {
		try {
			Locale.setDefault(Locale.US);
			boolean oj = System.getProperty("ONLINE_JUDGE") != null;
			Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("input.txt");
			Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("output.txt");
			in = new BufferedReader(reader);
			out = new PrintWriter(writer);
			solve();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public static void main(String[] args) {
		new D().run();

	}

}
