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.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class D {
    
    private StreamTokenizer in;
    private PrintWriter out;
    int max = 100005;
    int[] a = new int[max];
    int[] b = new int[max];
    
    public static void main (String[] args) throws IOException {
        new D().soilve();
    }
    
    int nextInt () throws IOException {
        in.nextToken();
        return (int)in.nval;
    }
    
    void soilve () throws IOException {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(new OutputStreamWriter(System.out));
        
        int n = nextInt();
        int m = nextInt();
        int c = nextInt();
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        int sum = 0;
        for (int i = 0; i < m; i++) {
            b[i] = nextInt();
            sum += b[i];
        }
        
        if (m == 1) {
            for (int i = 0; i < n; i++) {
                out.print(((a[i]+b[0]) % c)+ " ");
            }
            out.flush();
            return;
        }
        
        int sum1 = 0;
        int sum2 = 0;
        int k = n - m + 1;

        int h = 0, j = 0, l = 0;
        for (int i = 0; i <= n/2 - 1; i++) {
            if (h < k) {
                if (sum1 < sum) {
                    sum1 += b[j];
                }
                a[i] += sum1;
                a[i] %= c;
                
                if (sum2 < sum) {
                    sum2 += b[m - j - 1];
                }
                a[n - i - 1] += sum2;
                a[n - i - 1] %= c;
                
                h++;
                j++;
            } else {
                sum1 += b[j];
                sum1 -= b[l];
                a[i] += sum1;
                a[i] %= c;
                
                sum2 += b[m - j - 1];
                sum2 -= b[m - l - 1];
                a[n - i - 1] += sum2;
                a[n - i - 1] %= c;
                
                j++;
                l++;
            }
        }
        if (n % 2 == 1) {
            if (k >= m) {
                a[n / 2] += sum;
                a[n / 2] %= c;
            } else {
                sum1 += b[j];
                sum1 -= b[l];
                a[n / 2] += sum1;
                a[n / 2] %= c;
            }
        }
        
        for (int i = 0; i < n; i++) {
            out.print(a[i] + " ");
        }
        
        out.flush();
    }

}
