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;
struct Rohan {
  int x, y;
  int d, mask;
  string s;
};
struct compare {
  bool operator()(Rohan a, Rohan b) {
    if (a.d == b.d) {
      return (a.s > b.s);
    } else {
      return (a.d > b.d);
    }
  }
};
int distance(int x, int y, int a, int b) { return (abs(x - a) + abs(y - b)); }
const int mx = 55;
string a[mx];
int n, m, k, sx, sy, tx, ty;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
bool safe(int x, int y) {
  return (x >= 0 && x < n && y >= 0 && y < m && a[x][y] != 'S');
}
int countBit(int n) {
  int ans = 0;
  while (n > 0) {
    int d = n % 2;
    n = n / 2;
    ans = ans + (d == 1);
  }
  return ans;
}
string bfs() {
  Rohan r = {sx, sy, distance(sx, sy, tx, ty), 0, ""};
  priority_queue<Rohan, vector<Rohan>, compare> pq;
  set<pair<int, pair<int, int> > > s;
  pq.push(r);
  while (pq.empty() == false) {
    r = pq.top();
    pq.pop();
    if (a[r.x][r.y] == 'T') {
      return r.s;
    }
    if (s.find(make_pair(r.mask, make_pair(r.x, r.y))) != s.end()) {
      continue;
    }
    s.insert(make_pair(r.mask, make_pair(r.x, r.y)));
    for (int i = 0; i < 4; i++) {
      int row = r.x + dx[i];
      int col = r.y + dy[i];
      if (safe(row, col) == false) {
        continue;
      }
      if (row == tx && col == ty) {
        return r.s;
      }
      int next = r.mask | (1 << (a[row][col] - 97));
      if (countBit(next) <= k) {
        Rohan temp;
        temp.x = row;
        temp.y = col;
        temp.s = r.s + a[row][col];
        temp.d = distance(row, col, tx, ty) + r.s.length();
        temp.mask = next;
        pq.push(temp);
      }
    }
  }
  return "-1";
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (a[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  cout << bfs();
  return 0;
}