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.lang.*;
import java.math.*;
import java.io.*;
public class B{
    static PrintWriter w=new PrintWriter(System.out);
    public static void main(String [] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int c=sc.nextInt();
        int a[]=new int[n];
        int b[]=new int[m];
        for(int i=0;i<n;i++){
            a[i]=sc.nextInt();
        }
        for(int i=0;i<m;i++){
            b[i]=sc.nextInt();
        }
        if(n-m+1>=m){
            int fsum[]=new int[m];
            int bsum[]=new int[m];
            fsum[0]=b[0];
            bsum[m-1]=b[m-1];
            for(int i=1;i<m;i++){
                fsum[i]=(fsum[i-1]+b[i])%c;
            }
            for(int i=m-2;i>=0;i--){
                bsum[i]=(bsum[i+1]+b[i])%c;
            }
           
            int ans[]=new int[n];
            int l=0,r=m-1;
            
            for(int i=0;i<(int)Math.ceil((double)n/2);i++){
                //w.println(a[i]+" "+b[l]+" "+a[n-i-1]+" "+b[r]);
                ans[i]=(a[i]+fsum[l])%c;
                ans[n-i-1]=(a[n-i-1]+bsum[r])%c;
                if(l<m-1){
                    l++;
                    r--;
                }
            }
            for(int i=0;i<n;i++){
                w.print(ans[i]+" ");
            }
        }
        else{
            Deque<Integer> dq=new LinkedList<Integer>();
            int sum=0;
            for(int i=0;i<n;i++){
                dq.addLast(i);
                if(i<n-m+1){
                    sum+=b[i];
                }
                else if(i>=n-m+1&&i<m){
                    sum+=b[i];
                    sum-=b[dq.removeFirst()];
                }
                else if(i>=m){
                    sum-=b[dq.removeFirst()];
                    dq.removeLast();
                }
                
                System.out.print((sum+a[i])%c+" ");
            }
        }
        
        w.println();
        w.close();
    }
}