You are given an n × m table, consisting of characters «A», «G», «C», «T». Let's call a table nice, if every 2 × 2 square contains all four distinct characters. Your task is to find a nice table (also consisting of «A», «G», «C», «T»), that differs from the given table in the minimum number of characters.

Input

First line contains two positive integers n and m — number of rows and columns in the table you are given (2 ≤ n, m, n × m ≤ 300 000). Then, n lines describing the table follow. Each line contains exactly m characters «A», «G», «C», «T».

Output

Output n lines, m characters each. This table must be nice and differ from the input table in the minimum number of characters.

Examples

Input


2 2
AG
CT


Output


AG
CT


Input


3 5
AGCAG
AGCAG
AGCAG


Output


TGCAT
CATGC
TGCAT

Note

In the first sample, the table is already nice. In the second sample, you can change 9 elements to make the table nice.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 150005;
int mp[400], a[6], n, m, cnt, ans, f[6], type, t[2][4];
char ch[6], s[maxn];
vector<int> vc[maxn];
bool vis[6];
void calc(int o) {
  if (o == 1) {
    for (int i = 1; i <= n; i++) {
      int u = i & 1, now, temp = 0, v = INT_MAX;
      now = 1;
      for (int j = 1; j <= m; j++) {
        if (vc[i][j] != t[u][now]) temp++;
        now++;
        if (now > 2) now -= 2;
      }
      v = min(v, temp);
      now = 2;
      temp = 0;
      for (int j = 1; j <= m; j++) {
        if (vc[i][j] != t[u][now]) temp++;
        now++;
        if (now > 2) now -= 2;
      }
      v = min(v, temp);
      cnt += v;
    }
  } else {
    for (int i = 1; i <= m; i++) {
      int u = i & 1, now, temp = 0, v = INT_MAX;
      now = 1;
      for (int j = 1; j <= n; j++) {
        if (vc[j][i] != t[u][now]) temp++;
        now++;
        if (now > 2) now -= 2;
      }
      v = min(v, temp);
      now = 2;
      temp = 0;
      for (int j = 1; j <= n; j++) {
        if (vc[j][i] != t[u][now]) temp++;
        now++;
        if (now > 2) now -= 2;
      }
      v = min(v, temp);
      cnt += v;
    }
  }
}
void dfs(int step) {
  if (step == 4) {
    t[0][1] = a[1];
    t[0][2] = a[2];
    t[1][1] = a[3];
    t[1][2] = a[4];
    cnt = 0;
    calc(1);
    if (cnt < ans) {
      for (int i = 1; i <= 4; i++) f[i] = a[i];
      type = 1;
      ans = cnt;
    }
    cnt = 0;
    calc(2);
    if (cnt < ans) {
      for (int i = 1; i <= 4; i++) f[i] = a[i];
      type = 2;
      ans = cnt;
    }
    return;
  }
  for (int i = 1; i <= 4; i++) {
    if (vis[i]) continue;
    vis[i] = true;
    a[step + 1] = i;
    dfs(step + 1);
    vis[i] = false;
  }
}
int main() {
  mp['A'] = 1;
  mp['G'] = 2;
  mp['C'] = 3;
  mp['T'] = 4;
  ch[1] = 'A';
  ch[2] = 'G';
  ch[3] = 'C';
  ch[4] = 'T';
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) vc[i].push_back(0);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= m; j++) vc[i].push_back(mp[s[j]]);
  }
  ans = 1e9;
  dfs(0);
  t[0][1] = f[1];
  t[0][2] = f[2];
  t[1][1] = f[3];
  t[1][2] = f[4];
  if (type == 1) {
    for (int i = 1; i <= n; i++) {
      int u = i & 1, now, ca = 0, cb = 0;
      now = 1;
      for (int j = 1; j <= m; j++) {
        if (vc[i][j] != t[u][now]) ca++;
        now++;
        if (now > 2) now -= 2;
      }
      now = 2;
      for (int j = 1; j <= m; j++) {
        if (vc[i][j] != t[u][now]) cb++;
        now++;
        if (now > 2) now -= 2;
      }
      if (ca < cb)
        now = 1;
      else
        now = 2;
      for (int j = 1; j <= m; j++) {
        vc[i][j] = t[u][now];
        now++;
        if (now > 2) now -= 2;
      }
    }
  } else {
    for (int i = 1; i <= m; i++) {
      int u = i & 1, now, ca = 0, cb = 0;
      now = 1;
      for (int j = 1; j <= n; j++) {
        if (vc[j][i] != t[u][now]) ca++;
        now++;
        if (now > 2) now -= 2;
      }
      now = 2;
      for (int j = 1; j <= n; j++) {
        if (vc[j][i] != t[u][now]) cb++;
        now++;
        if (now > 2) now -= 2;
      }
      if (ca < cb)
        now = 1;
      else
        now = 2;
      for (int j = 1; j <= n; j++) {
        vc[j][i] = t[u][now];
        now++;
        if (now > 2) now -= 2;
      }
    }
  }
  for (int i = 1; i <= n; i++, puts(""))
    for (int j = 1; j <= m; j++) printf("%c", ch[vc[i][j]]);
  return 0;
}
