You already know that Valery's favorite sport is biathlon. Due to your help, he learned to shoot without missing, and his skills are unmatched at the shooting range. But now a smaller task is to be performed, he should learn to complete the path fastest.

The track's map is represented by a rectangle n × m in size divided into squares. Each square is marked with a lowercase Latin letter (which means the type of the plot), with the exception of the starting square (it is marked with a capital Latin letters S) and the terminating square (it is marked with a capital Latin letter T). The time of movement from one square to another is equal to 1 minute. The time of movement within the cell can be neglected. We can move from the cell only to side-adjacent ones, but it is forbidden to go beyond the map edges. Also the following restriction is imposed on the path: it is not allowed to visit more than k different types of squares (squares of one type can be visited an infinite number of times). Squares marked with S and T have no type, so they are not counted. But S must be visited exactly once — at the very beginning, and T must be visited exactly once — at the very end.

Your task is to find the path from the square S to the square T that takes minimum time. Among all shortest paths you should choose the lexicographically minimal one. When comparing paths you should lexicographically represent them as a sequence of characters, that is, of plot types.

Input

The first input line contains three integers n, m and k (1 ≤ n, m ≤ 50, n·m ≥ 2, 1 ≤ k ≤ 4). Then n lines contain the map. Each line has the length of exactly m characters and consists of lowercase Latin letters and characters S and T. It is guaranteed that the map contains exactly one character S and exactly one character T.

Pretest 12 is one of the maximal tests for this problem.

Output

If there is a path that satisfies the condition, print it as a sequence of letters — the plot types. Otherwise, print "-1" (without quotes). You shouldn't print the character S in the beginning and T in the end.

Note that this sequence may be empty. This case is present in pretests. You can just print nothing or print one "End of line"-character. Both will be accepted.

Examples

Input

5 3 2
Sba
ccc
aac
ccc
abT


Output

bcccc


Input

3 4 1
Sxyy
yxxx
yyyT


Output

xxxx


Input

1 3 3
TyS


Output

y


Input

1 4 1
SxyT


Output

-1
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int inf = 1e9 + 7;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
int sx, sy, tx, ty, dp[N][N];
char a[N][N];
string dfs(int x, int n, string s) {
  if (n) {
    string ret = "-1";
    for (int i = (x); i <= (25); i++) {
      string tmp = dfs(i + 1, n - 1, s + char(i + 'a'));
      if (tmp == "-1") continue;
      if (ret == "-1" || tmp.size() < ret.size() ||
          tmp.size() == ret.size() && tmp < ret)
        ret = tmp;
    }
    return ret;
  }
  vector<pair<int, int> > u, v;
  u.push_back(make_pair(tx, ty));
  memset(dp, 255, sizeof(dp));
  dp[tx][ty] = -2;
  while (!u.empty()) {
    for (int i = 0; i < s.length(); i++)
      for (int j = 0; j < u.size(); j++) {
        char c = s[i];
        int fr = u[j].first, fc = u[j].second;
        for (int k = (0); k <= (3); k++) {
          int nx = fr + dx[k], ny = fc + dy[k];
          if (a[nx][ny] != c || ~dp[nx][ny]) continue;
          dp[nx][ny] = k;
          v.push_back(make_pair(nx, ny));
          if (nx == sx && ny == sy) {
            string ret;
            while (1) {
              k = dp[nx -= dx[k]][ny -= dy[k]];
              if (nx == tx && ny == ty) break;
              ret += a[nx][ny];
            }
            return ret;
          }
        }
      }
    v.swap(u);
    v.clear();
  }
  return "-1";
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (1); i <= (n); i++) {
    scanf("%s", a[i] + 1);
    for (int j = (1); j <= (m); j++) {
      if (a[i][j] == 'S')
        sx = i, sy = j;
      else if (a[i][j] == 'T')
        tx = i, ty = j;
      if (a[i][j] < 'a' || a[i][j] > 'z') a[i][j] = '.';
    }
  }
  cout << dfs(0, k, ".") << endl;
  return 0;
}
