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(void) {
  int n, m, c;
  cin >> n >> m >> c;
  vector<int> a(n), b(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int j = 0; j < m; j++) cin >> b[j];
  vector<int> beg(n);
  beg.assign(n, 0);
  vector<int> start(n), end(n);
  start.assign(n, 0);
  end.assign(n, 0);
  int max;
  if (n >= 2 * m) {
    for (int i = 0; i < m; ++i) {
      beg[i] = i + 1;
      start[i] = 0;
      end[i] = start[i] + beg[i];
    }
    for (int i = m; i < n - m + 1; ++i) {
      beg[i] = m;
      start[i] = 0;
      end[i] = start[i] + beg[i];
    }
    for (int i = n - m + 1; i < n; ++i) {
      beg[i] = n - i;
      start[i] = start[i - 1] + 1;
      end[i] = start[i] + beg[i];
    }
  } else {
    for (int i = 0; i < n - m + 1; ++i) {
      beg[i] = i + 1;
      start[i] = 0;
      end[i] = start[i] + beg[i];
    }
    for (int i = n - m + 1; i < m; ++i) {
      beg[i] = n - m + 1;
      start[i] = start[i - 1] + 1;
      end[i] = start[i] + beg[i];
    }
    for (int i = m; i < n; ++i) {
      beg[i] = n - i;
      start[i] = start[i - 1] + 1;
      end[i] = start[i] + beg[i];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = start[i]; j < end[i]; ++j) a[i] = (a[i] + b[j]) % c;
    cout << a[i] << " ";
  }
  return 0;
}
