Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int c[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
bool f[30], vis[60][60];
char a[60][60];
int n, m, K, d[60][60];
string ans;
pair<int, int> S, T;
queue<pair<int, int> > Q;
vector<pair<int, int> > b;
void bfs() {
  memset(d, 63, sizeof(d));
  int inf = d[0][0];
  d[T.first][T.second] = 0;
  Q.push(T);
  while (!Q.empty()) {
    int x0 = Q.front().first, y0 = Q.front().second;
    Q.pop();
    for (int i = 0; i < 4; i++) {
      int x = x0 + c[i][0], y = y0 + c[i][1];
      if (x > 0 && x <= n && y > 0 && y <= m && d[x0][y0] + 1 < d[x][y] &&
          (a[x][y] == 'S' || f[a[x][y] - 'a'])) {
        d[x][y] = d[x0][y0] + 1;
        Q.push(make_pair(x, y));
      }
    }
  }
  if (d[S.first][S.second] == inf) return;
  string now;
  Q.push(S);
  memset(vis, 0, sizeof(vis));
  vis[S.first][S.second] = true;
  while (1) {
    char ch = 'z' + 1;
    b.clear();
    while (!Q.empty()) {
      int x0 = Q.front().first, y0 = Q.front().second;
      Q.pop();
      for (int i = 0; i < 4; i++) {
        int x = x0 + c[i][0], y = y0 + c[i][1];
        if (x > 0 && x <= n && y > 0 && y <= m && d[x0][y0] == d[x][y] + 1) {
          if (a[x][y] < ch) {
            b.clear();
            ch = a[x][y];
          }
          if (a[x][y] == ch && !vis[x][y]) b.push_back(make_pair(x, y));
          vis[x][y] = true;
        }
      }
    }
    if (b.size() == 0 || ch == 'T') break;
    now += ch;
    for (int i = 0; i < b.size(); i++) Q.push(b[i]);
  }
  if (now.size() < ans.size() || now.size() == ans.size() && now < ans)
    ans = now;
}
void dfs(int dep, int last) {
  if (dep > K) return;
  bfs();
  for (int i = last + 1; i < 26; i++) {
    f[i] = true;
    dfs(dep + 1, i);
    f[i] = false;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= n; i++) scanf("%s", a[i] + 1);
  for (int i = 1; i <= n * m; i++) ans += 'z';
  ans += 'z';
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == 'S') S = make_pair(i, j);
      if (a[i][j] == 'T') T = make_pair(i, j);
    }
  for (int i = 0; i < 26; i++) {
    f[i] = true;
    dfs(1, i);
    f[i] = false;
  }
  if (ans.size() == n * m + 1)
    printf("-1\n");
  else
    printf("%s\n", ans.c_str());
  return (0);
}