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;
long long Hash(char* s, int sz) {
  const int p = 31;
  long long hash = 0, p_pow = 1;
  for (size_t i = 0; i < sz; ++i) {
    hash += (s[i] - 'a' + 1) * p_pow;
    p_pow *= p;
  }
  return hash;
}
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int a[100042];
int key[100042];
int main() {
  int n, m, c;
  scanf("%d %d %d", &n, &m, &c);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) scanf("%d", &key[i]);
  int sum = key[0];
  int prev = 0;
  for (int i = 0; i < n; i++) {
    a[i] += sum;
    sum += key[i + 1];
    if (i - prev >= n - m) sum -= key[prev++];
    printf("%d ", a[i] % c);
  }
}
