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};
const int dy[4] = {0, 1, 0, -1};
struct nd {
  int x, y, h;
  string s;
  nd(){};
  nd(int x, int y, int h, string s) : x(x), y(y), h(h), s(s){};
  bool operator<(const nd &oth) const {
    if (oth.s.length() != s.length()) return s.length() > oth.s.length();
    return s > oth.s;
  }
};
char a[55][55];
int n, m, mx;
set<int> s[55][55];
int lst[55][55];
int bitcnt(int x) {
  int cnt = 0;
  for (int i = 0; i < 26; i++)
    if (x & (1 << i)) cnt++;
  return cnt;
}
bool dijk(int i, int j) {
  priority_queue<nd> pq;
  pq.push(nd(i, j, 0, ""));
  while (!pq.empty()) {
    nd cur = pq.top();
    pq.pop();
    for (int k = 0; k < 4; k++) {
      int x = cur.x + dx[k];
      int y = cur.y + dy[k];
      if (x < 0 || y < 0 || x >= n || y >= m) continue;
      if (a[x][y] == 'T') {
        cout << cur.s;
        return true;
      }
      int h = cur.h | (1 << (a[x][y] - 'a'));
      if (bitcnt(h) > mx || s[x][y].count(h)) continue;
      if (lst[x][y] && (lst[x][y] & h) == lst[x][y]) continue;
      lst[x][y] = h;
      s[x][y].insert(h);
      pq.push(nd(x, y, h, cur.s + a[x][y]));
    }
  }
  return false;
}
int main() {
  scanf("%d%d%d\n", &n, &m, &mx);
  for (int i = 0; i < n; i++) scanf("%s", a[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'S') {
        if (!dijk(i, j)) printf("-1");
        return 0;
      }
    }
  return 0;
}
