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 dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
struct state {
  int x, y, h;
  string s;
  state(int a, int b, int c, string d) : x(a), y(b), h(c), s(d) {}
  bool operator<(const state &b) const {
    return b.s.size() != s.size() ? s.size() > b.s.size() : s > b.s;
  }
};
char a[55][55];
int n, m, lim, last[55][55];
set<int> s[55][55];
int main() {
  cin >> n >> m >> lim;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j] == 'S') {
        priority_queue<state> q;
        q.push({i, j, 0, ""});
        while (!q.empty()) {
          auto cur = q.top();
          q.pop();
          for (int d = 0; d < 4; d++) {
            int x = cur.x + dx[d], y = cur.y + dy[d];
            if (x < 0 || y < 0 || x >= n || y >= m) continue;
            if (a[x][y] == 'T') {
              cout << cur.s << '\n';
              exit(0);
            }
            if (a[x][y] == 'S') continue;
            int h = cur.h | (1 << (a[x][y] - 'a'));
            if (__builtin_popcount(h) > lim || s[x][y].count(h)) continue;
            if (last[x][y] && (last[x][y] & h) == last[x][y]) continue;
            last[x][y] = h;
            s[x][y].insert(h);
            q.push({x, y, h, cur.s + a[x][y]});
          }
        }
        puts("-1");
        exit(0);
      }
}
