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 n, m, k, S, T;
char s[2555];
vector<int> adj[2555];
int val[2555], d[2555], p[2555];
string ans;
void check(int x) {
  string now;
  for (; d[x]; x = p[x]) now += s[x];
  reverse(now.begin(), now.end());
  if (now.length() < ans.length())
    ans = now;
  else if (now.length() == ans.length() && now < ans)
    ans = now;
}
bool cmp_val(int i, int j) { return val[i] < val[j]; }
void solve(int mask) {
  memset(d, -1, sizeof d);
  vector<int> u, v;
  d[S] = 0, val[S] = 0, u.push_back(S);
  while (!u.empty()) {
    v.clear();
    for (int i = 0; i < u.size(); i++) {
      int x = u[i];
      for (int j = 0; j < adj[x].size(); j++) {
        int y = adj[x][j];
        if (y == T) {
          check(x);
          return;
        }
        if (d[y] != -1 || !(mask >> (s[y] - 'a') & 1)) continue;
        d[y] = d[x] + 1, p[y] = x, val[y] = val[x] * 2555 + s[y],
        v.push_back(y);
      }
    }
    sort(v.begin(), v.end(), cmp_val);
    for (int i = 0, prv; i < v.size(); i++) {
      if (i == 0)
        prv = val[v[0]], val[v[0]] = 0;
      else {
        int tmp = val[v[i]];
        val[v[i]] = val[v[i - 1]] + (tmp != prv);
        prv = tmp;
      }
    }
    u = v;
  }
}
bool good(int i, int j) { return 0 <= i && i < n && 0 <= j && j < m; }
int node(int i, int j) { return i * m + j; }
bool cmp_s(int i, int j) { return s[i] < s[j]; }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", s + i * m);
  ans = string(3000, 'a');
  for (int i = 0; i < n * m; i++) {
    if (s[i] == 'S') S = i;
    if (s[i] == 'T') T = i;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (good(i + 1, j)) adj[node(i, j)].push_back(node(i + 1, j));
      if (good(i - 1, j)) adj[node(i, j)].push_back(node(i - 1, j));
      if (good(i, j + 1)) adj[node(i, j)].push_back(node(i, j + 1));
      if (good(i, j - 1)) adj[node(i, j)].push_back(node(i, j - 1));
    }
  for (int i = 0; i < n * m; i++) sort(adj[i].begin(), adj[i].end(), cmp_s);
  for (int i = 0; i < (1 << 26); i++)
    if (__builtin_popcount(i) == k) solve(i);
  if (ans.length() == 3000) ans = "-1";
  cout << ans << endl;
}
