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 n, m, c;
vector<int> a, b;
vector<pair<int, int> > arr;
void solve() {
  for (int i = 0; i < n; ++i) {
    int f = arr[i].first, l = arr[i].second;
    a[i] += b[l];
    if (f) {
      a[i] -= b[f - 1];
    }
  }
}
int main(int argc, char *argv[]) {
  scanf("%d%d%d", &n, &m, &c);
  a.resize(n);
  b.resize(m);
  arr.resize(n);
  for (int i = 0; i < n - m + 1; ++i) {
    arr[i].first = 0;
    arr[n - i - 1].second = m - 1;
  }
  for (int i = n - m + 1; i < n; ++i) {
    arr[i].first = i - n + m;
    arr[n - i - 1].second = m - (i - (n - m + 1) + 2);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d", &b[i]);
    if (i) {
      b[i] += b[i - 1];
    }
  }
  solve();
  for (int i = 0; i < n; ++i) {
    printf("%d ", a[i] % c);
  }
  return 0;
}
