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 nc, mc, c;
int n[100002];
int m[100002];
int cn[100002];
int Lowbit(int t) { return t & (t ^ (t - 1)); }
int Sum(int end) {
  int sum = 0;
  while (end > 0) {
    sum += cn[end];
    end -= Lowbit(end);
  }
  return sum;
}
int result(int j, int i) {
  int m1;
  int sum = 0;
  for (m1 = i; m1 <= j; m1++) {
    sum = sum + m[m1];
  }
  return sum;
}
void cinit(int pos) { cn[pos] = result(pos, pos - Lowbit(pos) + 1); }
int main() {
  memset(m, 0, m[0]);
  memset(n, 0, n[0]);
  memset(cn, 0, cn[0]);
  int i, j;
  scanf("%d%d%d", &nc, &mc, &c);
  for (i = 1; i <= nc; i++) {
    scanf("%d", &n[i]);
    n[i] = n[i] % c;
  }
  for (i = 1; i <= mc; i++) {
    scanf("%d", &m[i]);
    m[i] = m[i] % c;
    cinit(i);
  }
  if ((nc - mc + 1) >= (mc)) {
    for (i = 1; i <= mc; i++) {
      n[i] = n[i] + Sum(i);
    }
    for (i = mc + 1; i <= nc - mc + 1; i++) {
      n[i] = n[i] + Sum(mc);
    }
    for (i = nc - mc + 2, j = 1; i <= nc; i++, j++) {
      n[i] = n[i] + Sum(mc) - Sum(j);
    }
  } else {
    for (i = 1; i <= nc - mc + 1; i++) {
      n[i] = n[i] + Sum(i);
    }
    for (i = nc - mc + 2, j = 1; i <= mc; i++, j++) {
      n[i] = n[i] + Sum(i) - Sum(j);
    }
    for (i = mc + 1; i <= nc; i++, j++) {
      n[i] = n[i] + Sum(mc) - Sum(j);
    }
  }
  for (i = 1; i <= nc - 1; i++) {
    n[i] = n[i] % c;
    printf("%d ", n[i]);
  }
  printf("%d\n", n[nc] % c);
  return 0;
}
