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;
inline long long read() {
  long long S = 0;
  char ch = ' ';
  bool f = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    S = (S << 3) + (S << 1) + (ch - '0');
    ch = getchar();
  }
  return (f) ? (-S) : (S);
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 55;
int n, m, k, can[N][N];
char a[N][N];
int dx[] = {-1, 0, 0, 1}, dy[] = {0, -1, 1, 0};
struct node {
  int x, y, lim;
  string jyl;
  inline bool operator<(const node &tmp) const {
    return (tmp.jyl.size() != jyl.size()) ? (tmp.jyl.size() < jyl.size())
                                          : (tmp.jyl < jyl);
  }
};
priority_queue<node> que;
set<int> arr[N][N];
inline int js(int x) {
  int i, jyl = 0;
  for (i = 0; i <= 25; i++)
    if (x & (1 << i)) jyl++;
  return jyl;
}
int main() {
  int i, j, o, x, y, lim;
  n = read();
  m = read();
  k = read();
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (a[i][j] == 'S') {
        que.push((node){i, j, 0, ""});
        while (!que.empty()) {
          node now = que.top();
          que.pop();
          for (o = 0; o < 4; o++) {
            x = now.x + dx[o];
            y = now.y + dy[o];
            if (x < 0 || y < 0 || x >= n || y >= m || a[x][y] == 'S') continue;
            if (a[x][y] == 'T') {
              cout << now.jyl << endl;
              return 0;
            }
            lim = now.lim | (1 << (a[x][y] - 'a'));
            if (js(lim) > k || arr[x][y].count(lim)) continue;
            if (can[x][y] && (can[x][y] & lim) == can[x][y]) continue;
            can[x][y] = lim;
            arr[x][y].insert(lim);
            que.push((node){x, y, lim, now.jyl + a[x][y]});
          }
        }
        puts("-1");
      }
  return 0;
}
