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;
int n, m, k;
char board[3000][3000];
int si, sj, fi, fj;
bool found;
string ans;
bool outside(int x, int y) { return (x < 1 || x > n || y < 1 || y > m); }
int bits(int x) {
  if (x == 0) return 0;
  return bits(x / 2) + x % 2;
}
struct node {
  int x, y, mask;
  string way;
  bool operator<(const node &other) const {
    if (way.length() != other.way.length())
      return way.length() > other.way.length();
    return way > other.way;
  }
};
int sx, sy;
set<int> done[55][55];
priority_queue<node> qu;
vector<int> dx, dy;
int smask[55][55];
int main() {
  ios_base::sync_with_stdio(0);
  dx.push_back(0);
  dx.push_back(0);
  dx.push_back(1);
  dx.push_back(-1);
  dy.push_back(-1);
  dy.push_back(1);
  dy.push_back(0);
  dy.push_back(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    string st;
    cin >> st;
    for (int j = 1; j <= m; j++) {
      board[i][j] = st[j - 1];
      if (board[i][j] == 'S') {
        si = i;
        sj = j;
      }
      if (board[i][j] == 'T') {
        fi = i;
        fj = j;
      }
    }
  }
  node temp;
  temp.x = si;
  temp.y = sj;
  temp.mask = 0;
  temp.way = "";
  qu.push(temp);
  found = 0;
  while (qu.size()) {
    node cur = qu.top();
    qu.pop();
    int qx = cur.x;
    int qy = cur.y;
    if (found) break;
    for (int dir = 0; dir < 4; dir++) {
      int nx = qx + dx[dir];
      int ny = qy + dy[dir];
      if (outside(nx, ny)) continue;
      if (board[nx][ny] == 'S') continue;
      if (board[nx][ny] == 'T' && found == 0) {
        found = 1;
        ans = cur.way;
        break;
      }
      int nmask = (cur.mask | (1 << (board[nx][ny] - 'a')));
      if (bits(nmask) > k) continue;
      if (done[nx][ny].find(nmask) != done[nx][ny].end()) continue;
      if ((smask[nx][ny] & nmask) == smask[nx][ny] && smask[nx][ny] > 0)
        continue;
      smask[nx][ny] = nmask;
      done[nx][ny].insert(nmask);
      temp.x = nx;
      temp.y = ny;
      temp.mask = nmask;
      temp.way = cur.way + board[nx][ny];
      qu.push(temp);
    }
  }
  if (found == 0)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}
