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 nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
const int BUFSIZE = 1000000;
char buf[BUFSIZE + 1];
string nextString() {
  scanf("%s", buf);
  return buf;
}
struct Point {
  double x, y;
  Point() : x(0), y(0) {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator-(Point op) const { return Point(x - op.x, y - op.y); }
  Point operator+(Point op) const { return Point(x + op.x, y + op.y); }
  Point operator*(double op) const { return Point(x * op, y * op); }
  double operator*(Point op) const { return x * op.x + y * op.y; }
  double operator%(Point op) const { return x * op.y - y * op.x; }
  double length2() { return x * x + y * y; }
  double lengt() { return sqrt(length2()); }
};
Point nextPoint() {
  double x = nextDouble();
  double y = nextDouble();
  return Point(x, y);
}
int main() {
  int n = nextInt();
  int m = nextInt();
  int c = nextInt();
  vector<int> a;
  for (int i = 0; i < n; ++i) {
    a.push_back(nextInt());
  }
  vector<int> b;
  for (int i = 0; i < m; ++i) {
    b.push_back(nextInt());
  }
  for (int i = 0; i < (n - m + 1); ++i) {
    for (int j = 0; j < m; ++j) {
      a[j + i] += b[j];
      a[j + i] %= c;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << a[i] << " ";
  }
  cout << endl;
  return 0;
}
