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;
bool cmp(string a, string b) {
  if (a.length() != b.length()) return a.length() < b.length();
  return a < b;
}
struct node {
  int x, y, mask;
  string path;
  node() {}
  node(int x, int y, int m, string p) : x(x), y(y), mask(m), path(p) {}
  bool operator<(const node &p) const { return cmp(p.path, path); }
};
char s[55][55];
set<int> vis[55][55];
int f[55][55];
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
string bfs(int x, int y, int k) {
  string res(55 * 55, 'a');
  priority_queue<node> q;
  q.push(node(x, y, 0, ""));
  vis[x][y].insert(0);
  f[x][y] = 0;
  while (!q.empty()) {
    node u = q.top();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int x = u.x + dx[i], y = u.y + dy[i];
      if (!s[x][y] || s[x][y] == 'S') continue;
      if (s[x][y] == 'T') {
        if (cmp(u.path, res)) res = u.path;
        break;
      }
      int z = u.mask | (1 << (s[x][y] - 'a'));
      if (__builtin_popcount(z) > k) continue;
      if (vis[x][y].count(z)) continue;
      if (f[x][y] && (f[x][y] & z) == f[x][y]) continue;
      vis[x][y].insert(z);
      f[x][y] = z;
      q.push(node(x, y, z, u.path + s[x][y]));
    }
  }
  if (res.length() == 55 * 55) res = "-1";
  return res;
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (s[i][j] == 'S') {
        cout << bfs(i, j, k) << endl;
        return 0;
      }
    }
}
