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;
struct node {
  int x, y, msk;
  string s;
  node(int x1, int x2, int x3, string x4) { x = x1, y = x2, msk = x3, s = x4; }
  bool operator<(const node &b) const {
    if (s.size() != b.s.size()) return s.size() > b.s.size();
    return s > b.s;
  }
};
int gx[5] = {0, 1, 0, -1, 0}, gy[5] = {0, 0, 1, 0, -1};
char s[51][51];
set<int> st[51][51];
int las[51][51];
int count(int x) {
  int ans = 0;
  while (x) ans += x & 1, x >>= 1;
  return ans;
}
signed main() {
  int x = 0, y = 0;
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) cin >> s[i] + 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (s[i][j] == 'S') x = i, y = j;
  priority_queue<node> q;
  q.push(node(x, y, 0, ""));
  while (!q.empty()) {
    node u = q.top();
    q.pop();
    for (int i = 1; i <= 4; i++) {
      int tox = u.x + gx[i];
      int toy = u.y + gy[i];
      if (tox > n || toy > m || tox <= 0 || toy <= 0) continue;
      if (s[tox][toy] == 'T') {
        cout << u.s;
        return 0;
      }
      int mk = u.msk | (1 << (s[tox][toy] - 'a'));
      if (count(mk) > k) continue;
      if (las[tox][toy] && ((las[tox][toy] & mk) == las[tox][toy])) continue;
      las[tox][toy] = mk;
      if (st[tox][toy].count(mk)) continue;
      st[tox][toy].insert(mk);
      string tmp = u.s;
      q.push(node(tox, toy, mk, u.s + s[tox][toy]));
    }
  }
  printf("-1");
  return 0;
}
