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;
int main() {
  int inf = 10000;
  string best = "";
  int min = 10000;
  int n, m, k;
  char a[50][51];
  int start = 0;
  int stop = 0;
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  string ret[2500];
  bool hasValue[2500];
  bool visted[2500];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'S')
        start = i * m + j;
      else if (a[i][j] == 'T')
        stop = i * m + j;
    }
  }
  int c[4];
  for (int i = 0; i < k; i++) {
    c[i] = i;
  }
  while (true) {
    bool b[256];
    memset(b, 0, sizeof(b));
    for (int i = 0; i < k; i++) {
      b[c[i] + 'a'] = true;
    }
    b['S'] = b['T'] = true;
    queue<int> q;
    q.push(stop);
    memset(hasValue, 0, sizeof(hasValue));
    memset(visted, 0, sizeof(visted));
    ret[stop] = string("");
    hasValue[stop] = true;
    visted[stop] = true;
    int delta[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    while (q.size() > 0) {
      int v = q.front();
      q.pop();
      int x = v / m;
      int y = v % m;
      string val = string(1, a[x][y]) + ret[v];
      if (a[x][y] == 'T') val = string("");
      if (v == start) {
        if (ret[v].size() < min) {
          best = ret[v];
          min = ret[v].size();
        } else if (ret[v].size() == min) {
          if (ret[v] < best) best = ret[v];
        }
        break;
      }
      for (int i = 0; i < 4; i++) {
        int nx = x + delta[i][0];
        int ny = y + delta[i][1];
        int cm = nx * m + ny;
        if (0 <= nx && nx < n && 0 <= ny && ny < m && b[a[nx][ny]]) {
          if (!hasValue[cm] || val < ret[cm]) {
            ret[cm] = val;
            hasValue[cm] = true;
          }
          if (!visted[cm]) {
            q.push(cm);
            visted[cm] = true;
          }
        }
      }
    }
    c[k - 1]++;
    int z = k - 1;
    while (c[z] >= 26 - ((k - 1) - z) && z >= 1) {
      c[z - 1]++;
      for (int i = z; i < k; i++) {
        c[i] = c[z - 1] + i - z + 1;
      }
      z--;
    }
    if (c[0] >= 26 - k + 1) {
      break;
    }
  }
  if (min == 10000) {
    cout << -1 << endl;
  } else {
    cout << best << endl;
  }
}
