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 MAXn = 50 + 10, MAXk = 5;
char ch[MAXn][MAXn], sal[MAXk];
string ans;
int n, m, k, dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
string dis[MAXn][MAXn];
bool flag[MAXn][MAXn];
pair<int, int> S, T;
bool cmp(string &s1, string &s2) {
  if (s1.empty() || s2.empty()) return s2.empty();
  if (s1.size() != s2.size()) return s1.size() < s2.size();
  return s1 < s2;
}
bool isok(int x, int y) {
  bool smt = false;
  for (int i = 1; i <= k; i++)
    if (ch[x][y] == sal[i]) smt = true;
  return (smt || ch[x][y] == 'T') && min(x, y) >= 0 && x < n && y < m;
}
void bfs() {
  for (int i = 0; i < MAXn; i++)
    for (int j = 0; j < MAXn; j++) {
      dis[i][j].clear();
      flag[i][j] = false;
    }
  vector<pair<int, int> > vec;
  vec.push_back(S);
  flag[S.first][S.second] = true;
  for (int i = 0; i < vec.size();)
    for (int j = vec.size(); i < j; i++) {
      int x = vec[i].first, y = vec[i].second;
      string hp = dis[x][y];
      if (pair<int, int>(x, y) == T) {
        if (cmp(hp, ans)) ans = hp;
        return;
      }
      hp += ch[x][y];
      for (int z = 0; z < 4; z++) {
        int tmpx = x + dx[z], tmpy = y + dy[z];
        if (isok(tmpx, tmpy) && cmp(hp, dis[tmpx][tmpy])) {
          dis[tmpx][tmpy] = hp;
          if (!flag[tmpx][tmpy]) {
            flag[tmpx][tmpy] = true;
            vec.push_back(pair<int, int>(tmpx, tmpy));
          }
        }
      }
    }
}
void makeit(int a, int b) {
  if (b == 0) {
    bfs();
    return;
  }
  for (int i = 0; i < a; i++) {
    sal[b] = 'a' + i;
    makeit(i, b - 1);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> ch[i][j];
      if (ch[i][j] == 'S') S = pair<int, int>(i, j);
      if (ch[i][j] == 'T') T = pair<int, int>(i, j);
    }
  makeit(26, k);
  if (ans.empty()) return cout << -1, 0;
  for (int i = 1; i < ans.size(); i++) cout << ans[i];
}
