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;
string res;
int c[26], w[26];
bool vis[26];
int f[60][60];
char g[60][60];
int n, m, k, s1, s2, tot, step;
int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
struct In {
  int l;
  int r;
  string ss;
} s[3000];
void bfs() {
  memset(f, -1, sizeof(f));
  int i, j, first = 0, last = 1, len = 1, move = 1, ll;
  s[0].l = s1;
  s[0].r = s2;
  s[0].ss = "";
  f[s1][s2] = 0;
  string a;
  bool z = false;
  while (first < last) {
    for (i = first; i < last; i++) {
      for (j = 0; j < 4; j++) {
        int p, q;
        p = s[i].l + way[j][0];
        q = s[i].r + way[j][1];
        if (p >= 0 && p < n && q >= 0 && q < m) {
          if (f[p][q] != -1 && f[p][q] < move) continue;
          a = s[i].ss;
          if (g[p][q] == 'T') {
            if (res == "") {
              res = a;
              step = move;
            } else {
              int x1 = a.length(), x2 = res.length();
              if (x1 < x2 || (x1 == x2 && a < res)) {
                res = a;
                if (move < step) step = move;
              }
            }
            z = true;
          } else if (!z) {
            bool zz = false;
            for (ll = 0; ll < tot && ll < k; ll++) {
              if (w[ll] == g[p][q] - 'a') {
                zz = true;
                break;
              }
            }
            if (zz) {
              a += g[p][q];
              for (ll = last; ll < len; ll++)
                if (s[ll].l == p && s[ll].r == q) {
                  if (s[ll].ss > a) s[ll].ss = a;
                  break;
                }
              if (ll == len) {
                s[len].l = p;
                s[len].r = q;
                s[len].ss = a;
                len++;
                f[p][q] = move;
              }
            }
          }
        }
      }
    }
    if (z) break;
    first = last;
    last = len;
    move++;
    if (move > step) break;
  }
}
void dfs(int x, int y) {
  if (x == tot) return;
  w[y] = c[x];
  if (y == k - 1) {
    bfs();
  } else
    dfs(x + 1, y + 1);
  dfs(x + 1, y);
}
int main() {
  while (cin >> n >> m >> k) {
    int i, j;
    tot = 0;
    memset(vis, 0, sizeof(vis));
    for (i = 0; i < n; i++) {
      cin >> g[i];
      for (j = 0; j < m; j++) {
        if (g[i][j] == 'S') {
          s1 = i;
          s2 = j;
        } else if (g[i][j] >= 'a' && g[i][j] <= 'z') {
          if (!vis[g[i][j] - 'a']) {
            c[tot++] = g[i][j] - 'a';
            vis[g[i][j] - 'a'] = 1;
          }
        }
      }
    }
    step = 3000;
    if (tot <= k) {
      for (i = 0; i < tot; i++) w[i] = c[i];
      bfs();
    } else
      dfs(0, 0);
    if (step == 3000)
      cout << -1 << endl;
    else
      cout << res << endl;
  }
  return 0;
}
