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 long long MOD = 1e9 + 7;
const int INF = 2e9;
const long long INF64 = 9e18;
const long double EPS = 1e-12;
const long double PI = 3.1415926535897932384626433832795028841;
const long long MD = 925262732;
const long long T = 634521;
const int N = 200010;
const int M = 201;
int n, m, k, c[5], mini = INF, d[51][51];
bool mark[51][51];
string ans, a[M];
pair<int, int> s, f;
queue<pair<int, int> > q;
vector<pair<int, int> > qq[3];
bool get(int first, int second) {
  if (!(first >= 0 && second >= 0 && first < n && second < m)) return 0;
  bool f = 0;
  for (int i = 0; i < int(k); i++)
    if (a[first][second] == c[i]) f = 1;
  if (a[first][second] == 'S') f = 1;
  return f;
}
void bfs() {
  for (int i = 0; i < int(n); i++)
    for (int j = 0; j < int(m); j++) {
      d[i][j] = INF;
      mark[i][j] = 0;
    }
  d[f.first][f.second] = 0;
  q.push(f);
  while (!q.empty()) {
    pair<int, int> v = q.front();
    q.pop();
    for (int i = -1; i <= 1; i++)
      for (int j = -1; j <= 1; j++)
        if (abs(i) + abs(j) == 1)
          if (get(v.first + i, v.second + j) &&
              d[v.first + i][v.second + j] > d[v.first][v.second] + 1) {
            q.push(make_pair(v.first + i, v.second + j));
            d[v.first + i][v.second + j] = d[v.first][v.second] + 1;
          }
  }
  int sum = d[s.first][s.second];
  int ss = sum;
  if (sum == INF) return;
  if (sum == 1) exit(0);
  string s1 = "";
  int first = s.first, second = s.second;
  qq[0].clear();
  qq[1].clear();
  qq[sum % 2].push_back(s);
  while (sum > 1) {
    int c = 'z';
    int t = sum % 2, tt = (sum - 1) % 2;
    qq[tt].clear();
    for (int u = 0; u < int(int((qq[t]).size())); u++)
      for (int i = -1; i <= 1; i++)
        for (int j = -1; j <= 1; j++)
          if (abs(i) + abs(j) == 1) {
            int xx = qq[t][u].first + i, yy = qq[t][u].second + j;
            if (get(xx, yy) && d[xx][yy] == sum - 1 && a[xx][yy] <= c)
              c = a[xx][yy];
          }
    for (int u = 0; u < int(int((qq[t]).size())); u++)
      for (int i = -1; i <= 1; i++)
        for (int j = -1; j <= 1; j++)
          if (abs(i) + abs(j) == 1) {
            int xx = qq[t][u].first + i, yy = qq[t][u].second + j;
            if (get(xx, yy) && d[xx][yy] == sum - 1 && a[xx][yy] == c &&
                !mark[xx][yy]) {
              qq[tt].push_back(make_pair(xx, yy));
              mark[xx][yy] = 1;
            }
          }
    if (c != 'S') s1 = s1 + (char)c;
    sum--;
  }
  if (ss < mini) {
    mini = ss;
    ans = s1;
  } else if (ss == mini)
    ans = min(ans, s1);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < int(n); i++) {
    cin >> a[i];
    for (int j = 0; j < int(m); j++)
      if (a[i][j] == 'S')
        s = make_pair(i, j);
      else if (a[i][j] == 'T')
        f = make_pair(i, j);
  }
  for (c[0] = 'a'; c[0] <= 'z'; c[0]++)
    for (c[1] = c[0]; c[1] <= 'z'; c[1]++)
      for (c[2] = c[1]; c[2] <= 'z'; c[2]++)
        for (c[3] = c[2]; c[3] <= 'z'; c[3]++) bfs();
  if (mini == INF)
    cout << -1;
  else
    cout << ans;
  return 0;
}