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. 
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, c;
  int i, j;
  scanf("%d%d%d", &n, &m, &c);
  int *a = (int *)malloc((n + 1) * sizeof(int));
  int *b = (int *)malloc((n + 1) * sizeof(int));
  int *sb = (int *)malloc((n + 1) * sizeof(int));
  for (i = 0; i <= n - 1; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i <= m - 1; i++) {
    scanf("%d", &b[i]);
  }
  sb[0] = b[0];
  for (i = 1; i <= m - 1; i++) {
    sb[i] = (1LL * sb[i - 1] + b[i]) % c;
  }
  for (i = m; i <= n - 1; i++) {
    sb[i] = sb[i - 1];
  }
  int l = n - m;
  for (i = 0; i <= l; i++) {
    a[i] = (1LL * a[i] + sb[i]) % c;
  }
  int st = sb[l];
  j = 0;
  int k = l + 1;
  int temp;
  for (i = l + 1; i <= n - 1; i++) {
    if (k < m) {
      temp = (1LL * st - b[j] + b[k] + c) % c;
    } else {
      temp = (1LL * st - b[j] + c) % c;
    }
    a[i] = (1LL * a[i] + (1LL * st - b[j++] + (k < m ? b[k++] : 0)) + c) % c;
    st = temp;
  }
  for (i = 0; i <= n - 1; i++) {
    printf("%d ", a[i]);
  }
  return 0;
}
