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;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int gcd(int a, int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline int And(int mask, int bit) { return mask & (1 << bit); }
inline int Or(int mask, int bit) { return mask | (1 << bit); }
inline int Xor(int mask, int bit) { return mask & (~(1 << bit)); }
int a[100010], b[100010];
int sum[100010];
int main() {
  int n, m, c;
  scanf("%d %d %d", &n, &m, &c);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sum[0] = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    sum[i] = sum[i - 1] + b[i];
  }
  if (n == m) {
    for (int i = 1; i <= n; i++) a[i] = (a[i] + b[i]) % c;
  } else {
    for (int i = 1; i <= n - m + 1; i++)
      if (i <= m)
        a[i] = (a[i] + sum[i]) % c;
      else
        a[i] = (a[i] + sum[m]) % c;
    int x = 1;
    int y;
    if (n - m + 2 <= m)
      y = n - m + 2;
    else
      y = m;
    for (int i = n - m + 2; i <= n; i++) {
      int diff = sum[y] - sum[x];
      a[i] = (a[i] + diff) % c;
      x++;
      if (y + 1 <= m) y++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%d", a[i]);
  }
  printf("\n");
}
