Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int R, C, KK;
char ssG[N][N];
int stx, sty;
struct nodexx {
  string str;
  int x, y, val;
  nodexx(string str00, int x00, int y00, int val00) {
    str = str00, x = x00, y = y00, val = val00;
  }
  bool operator<(const nodexx t) const {
    int lenaa = str.size(), lenbb = t.str.size();
    if (lenaa - lenbb) return lenaa > lenbb;
    return str > t.str;
  }
};
priority_queue<nodexx> QQ;
set<pair<pair<int, int>, int> > SET;
int has[N][N];
int dirc[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
bool Judge(int x) {
  int cnt = 0;
  while (x > 0) {
    cnt += x % 2;
    x >>= 1;
  }
  return cnt <= KK;
}
string BFS() {
  SET.clear();
  memset(has, 0, sizeof has);
  while (QQ.size()) QQ.pop();
  QQ.push(nodexx("", stx, sty, 0));
  SET.insert(make_pair(make_pair(stx, sty), 0));
  int e, r, c, t;
  while (QQ.size()) {
    nodexx XA = QQ.top();
    QQ.pop();
    for (e = 0; e < 4; e++) {
      r = XA.x + dirc[e][0], c = XA.y + dirc[e][1];
      if (r >= 0 && r < R && c >= 0 && c < C && ssG[r][c] != 'S') {
        if (ssG[r][c] == 'T') return XA.str;
        t = XA.val | (1 << (ssG[r][c] - 'a'));
        if (Judge(t) && SET.count(make_pair(make_pair(r, c), t)) == 0) {
          if (has[r][c] && (has[r][c] & t) == has[r][c]) continue;
          has[r][c] = t;
          SET.insert(make_pair(make_pair(r, c), t));
          QQ.push(nodexx(XA.str + ssG[r][c], r, c, t));
        }
      }
    }
  }
  return "-1";
}
void cyh() { cout << BFS() << endl; }
int main() {
  int e, f;
  while (cin >> R >> C >> KK) {
    for (e = 0; e < R; e++) {
      scanf("%s", ssG[e]);
      for (f = 0; f < C; f++)
        if (ssG[e][f] == 'S') stx = e, sty = f;
    }
    cyh();
  }
  return 0;
}