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;
const int INF = 1000000000;
const long double pi = 3.1415926535897932384626433832795;
const long double EPS = 1e-9;
int d[100500], a[100500], b[100500];
void solve() {
  int n, m, c;
  cin >> n >> m >> c;
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int j = 0; j < m; ++j) scanf("%d", &b[j]);
  if (n >= 2 * m) {
    d[0] = b[0];
    for (int i = 1; i < m; i++) d[i] = d[i - 1] + b[i];
    for (int i = m; i <= n - m; i++) d[i] = d[i - 1];
    d[n - 1] = b[m - 1];
    for (int i = n - 2; i > n - m; i--)
      d[i] = d[i + 1] + b[m - 1 - (n - 1 - i)];
  } else {
    d[0] = b[0];
    for (int i = 1; i <= min(m - 1, n - m); i++) d[i] = d[i - 1] + b[i];
    for (int i = min(m - 1, n - m) + 1; i < n - min(m - 1, n - m) - 1; i++)
      d[i] = d[i - 1] + b[i] - b[i - (n - m + 1)];
    d[n - 1] = b[m - 1];
    for (int i = n - 2; i >= n - min(m - 1, n - m) - 1; i--)
      d[i] = d[i + 1] + b[m - 1 - (n - 1 - i)];
  }
  for (int i = 0; i < n; ++i) {
    printf("%d ", (a[i] + d[i]) % c);
  }
  cout << endl;
}
int main() {
  solve();
  return 0;
}
