You have an n × m rectangle table, its cells are not initially painted. Your task is to paint all cells of the table. The resulting picture should be a tiling of the table with squares. More formally:

  * each cell must be painted some color (the colors are marked by uppercase Latin letters); 
  * we will assume that two cells of the table are connected if they are of the same color and share a side; each connected region of the table must form a square. 



Given n and m, find lexicographically minimum coloring of the table that meets the described properties.

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 100).

Output

Print lexicographically minimum coloring of the table that meets the described conditions. 

One coloring (let's call it X) is considered lexicographically less than the other one (let's call it Y), if:

  * consider all the table cells from left to right and from top to bottom (first, the first cell in the first row, then the second cell in the first row and so on); 
  * let's find in this order the first cell that has distinct colors in two colorings; 
  * the letter that marks the color of the cell in X, goes alphabetically before the letter that marks the color of the cell in Y. 

Examples

Input

1 3


Output

ABA


Input

2 2


Output

AA
AA


Input

3 4


Output

AAAB
AAAC
AAAB
#include <bits/stdc++.h>
using namespace std;
inline long long Min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
inline long long Max(long long x, long long y) {
  if (x > y) return x;
  return y;
}
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
inline void Swap(long long &x, long long &y) {
  long long Temp = x;
  x = y;
  y = Temp;
}
bool Rotated;
long long Tian[105][105];
int main() {
  long long n, m;
  read(n);
  read(m);
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= m; ++j) {
      if (Tian[i][j]) continue;
      if (Tian[i][j - 1] != 1 && Tian[i - 1][j] != 1 && Tian[i][j + 1] != 1) {
        long long Hen = j, Zon = i;
        while (Hen < m && !Tian[i][Hen + 1] && Tian[i][Hen + 2] != 1 &&
               Tian[i - 1][Hen + 1] != 1)
          Hen++;
        while (Zon < n && !Tian[Zon + 1][j] && Tian[Zon + 2][j] != 1 &&
               Tian[Zon + 1][j - 1] != 1)
          Zon++;
        long long Bc = Min(Hen - j + 1, Zon - i + 1);
        for (register int k = i; k < i + Bc; ++k)
          for (register int l = j; l < j + Bc; ++l) Tian[k][l] = 1;
      } else {
        bool Vis2 = 0, Vis3 = 0;
        if (Tian[i - 1][j] == 2 || Tian[i][j - 1] == 2 || Tian[i][j + 1] == 2)
          Vis2 = 1;
        if (Tian[i - 1][j] == 3 || Tian[i][j - 1] == 3 || Tian[i][j + 1] == 3)
          Vis3 = 1;
        if (!Vis2) {
          long long Hen = j, Zon = i, Lia = j;
          while (Lia < m && Tian[i - 1][Lia + 1] == 1) Lia++;
          while (Hen < m && !Tian[i][Hen + 1]) Hen++;
          while (Zon < n && !Tian[Zon + 1][j]) Zon++;
          long long Bc = Min(Lia - j + 1, Min(Hen - j + 1, Zon - i + 1));
          for (register int k = i; k < i + Bc; ++k)
            for (register int l = j; l < j + Bc; ++l) Tian[k][l] = 2;
        } else if (!Vis3)
          Tian[i][j] = 3;
        else
          Tian[i][j] = 4;
      }
    }
  for (register int i = 1; i <= n; ++i, putchar('\n'))
    for (register int j = 1; j <= m; ++j) printf("%c", 'A' + Tian[i][j] - 1);
  return 0;
}
