You are given a cube of size k × k × k, which consists of unit cubes. Two unit cubes are considered neighbouring, if they have common face.

Your task is to paint each of k3 unit cubes one of two colours (black or white), so that the following conditions must be satisfied:

  * each white cube has exactly 2 neighbouring cubes of white color; 
  * each black cube has exactly 2 neighbouring cubes of black color. 

Input

The first line contains integer k (1 ≤ k ≤ 100), which is size of the cube.

Output

Print -1 if there is no solution. Otherwise, print the required painting of the cube consequently by layers. Print a k × k matrix in the first k lines, showing how the first layer of the cube should be painted. In the following k lines print a k × k matrix — the way the second layer should be painted. And so on to the last k-th layer. Note that orientation of the cube in the space does not matter.

Mark a white unit cube with symbol "w" and a black one with "b". Use the format of output data, given in the test samples. You may print extra empty lines, they will be ignored.

Examples

Input

1


Output

-1


Input

2


Output

bb
ww

bb
ww
#include <bits/stdc++.h>
using namespace std;
void rwFile() {
  freopen(
      "task"
      ".in",
      "r", stdin);
  freopen(
      "task"
      ".out",
      "w", stdout);
}
const long long MOD = 1e9 + 7;
const int maxN = 1e6 + 10;
const long long Inf = 1e18;
long long Pow(long long b, long long e) {
  if (e == 0) return 1;
  long long x = Pow(b, e / 2);
  if (e % 2 == 0) return x * x;
  return x * x * b;
}
long long modPow(long long b, long long e = MOD - 2) {
  if (e == 0) return 1;
  long long x = modPow(b, e / 2);
  if (e % 2 == 0) return x * x % MOD;
  return (x * x % MOD) * b % MOD;
}
int mobius(int a) {
  if (a == 1) return 1;
  int c = 0;
  for (int i = 2; i <= a; i++) {
    if (a % i == 0) {
      int p = 0;
      while (a % i == 0) {
        a /= i;
        p++;
      }
      if (p > 1) return 0;
      c++;
    }
  }
  if (c % 2 == 0) return 1;
  return -1;
}
long long preSum2(long long a) {
  long long res = a * (a + 1) % MOD;
  res = res * (2 * a + 1) % MOD;
  res = res * ((MOD + 1) / 6) % MOD;
  return res;
}
long long preSum3(long long a) {
  long long res = a * (a + 1) % MOD;
  res = res * ((MOD + 1) / 2) % MOD;
  res = res * res % MOD;
  return res;
}
long long G(long long n, int e) {
  if (e == 2) {
    long long res = 0;
    long long start = 2;
    while (start * start <= n) {
      long long tmp = n / start / start;
      long long endd = n / tmp;
      endd = sqrtl(endd);
      long long Sum = preSum2(endd) - preSum2(start - 1);
      Sum = (Sum + MOD) % MOD;
      res += (tmp % MOD) * Sum % MOD;
      start = endd + 1;
    }
    return res;
  }
  if (e == 3) {
    long long res = 0;
    long long start = 2;
    while (start * start * start <= n) {
      long long tmp = n / start / start / start;
      long long endd = n / tmp;
      endd = cbrtl(endd);
      long long Sum = preSum3(endd) - preSum3(start - 1);
      Sum = (Sum + MOD) % MOD;
      res += (tmp % MOD) * Sum % MOD;
      start = endd + 1;
    }
    return res;
  }
  int mob = mobius(e);
  if (!mob) return 0;
  long long Max = (long long)pow(10., 18. / e);
  long long res = 0;
  for (long long i = 2; i <= Max; i++) {
    long long tmp = Pow(i, e);
    if (tmp > n) break;
    long long temp = (n / tmp) % MOD;
    temp = temp * (tmp % MOD) % MOD;
    res = (res + temp) % MOD;
  }
  if (mob == 1) return (MOD - res) % MOD;
  return res;
}
void PPDIV() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long res = n;
    for (int i = 2; i < 60; i++) {
      if (Pow(2, i) > n) break;
      res += G(n, i);
      res %= MOD;
    }
    cout << res << '\n';
  }
}
void solveA_BlackAndWhiteCube() {
  int k;
  cin >> k;
  if (k % 2 == 1) {
    cout << -1;
    return;
  }
  char one = 'b', two = 'w';
  for (int i = 0; i < k; i++) {
    string res[k];
    char a = one, b = two;
    for (int j = 0; j < k; j++) {
      if (j == 0) {
        for (int l = 0; l < k; l++) res[j] += a;
        continue;
      }
      if (j >= k / 2) {
        res[j] = res[k - 1 - j];
        continue;
      }
      res[j] = res[j - 1];
      for (int l = j; l < k - j; l++) {
        res[j][l] = b;
      }
      swap(a, b);
    }
    for (string str : res) {
      cout << str << '\n';
    }
    cout << '\n';
    swap(one, two);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solveA_BlackAndWhiteCube();
}
