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;
string mat[50];
bool allow[26];
bool isAllow(char c) { return c == 'T' || c == 'S' || allow[c - 'a']; }
int di[] = {0, 0, 1, -1};
int dj[] = {1, -1, 0, 0};
int dist[50][50];
pair<int, int> pa[50][50];
int n, m;
bool in(int a, int b) { return 0 <= a && a < n && 0 <= b && b < m; }
void bfs(int x, int y) {
  pair<int, int> orig = make_pair(x, y);
  queue<pair<int, int> > cola;
  memset(dist, -1, sizeof(dist));
  int vis[n][m];
  memset(vis, 0, sizeof(vis));
  cola.push(orig);
  vis[x][y] = true;
  dist[x][y] = 0;
  while (!cola.empty()) {
    pair<int, int> cu = cola.front();
    cola.pop();
    int xx = cu.first, yy = cu.second;
    for (int k = 0; k < 4; k++) {
      int nx = xx + di[k], ny = yy + dj[k];
      if (in(nx, ny) && isAllow(mat[nx][ny])) {
        if (!vis[nx][ny]) {
          dist[nx][ny] = dist[xx][yy] + 1;
          pa[nx][ny] = cu;
          vis[nx][ny] = true;
          cola.push(make_pair(nx, ny));
        } else if (dist[nx][ny] == dist[xx][yy] + 1) {
          pair<int, int> p1 = cu, p2 = pa[nx][ny];
          while (p1 != orig &&
                 mat[p1.first][p1.second] == mat[p2.first][p2.second]) {
            p1 = pa[p1.first][p1.second];
            p2 = pa[p2.first][p2.second];
          }
          if (p1 != orig &&
              mat[p1.first][p1.second] < mat[p2.first][p2.second]) {
            pa[nx][ny] = cu;
          }
        }
      }
    }
  }
}
int main() {
  int kk;
  cin >> n >> m >> kk;
  int xS, yS, xT, yT;
  for (int i = 0; i < n; i++) {
    cin >> mat[i];
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (mat[i][j] == 'S') {
        xS = i;
        yS = j;
      }
      if (mat[i][j] == 'T') {
        xT = i;
        yT = j;
      }
    }
  pair<int, int> final = make_pair(xT, yT);
  string best;
  int largo = 1000000000;
  memset(allow, 0, sizeof(allow));
  bool first = true;
  for (int i = 0; i < 26; i++)
    for (int j = i; j < 26; j++)
      for (int k = j; k < 26; k++)
        for (int h = k; h < 26; h++) {
          if (kk < 4) h = i;
          if (kk < 3) k = i;
          if (kk < 2) j = i;
          allow[i] = allow[j] = allow[k] = allow[h] = true;
          bfs(xT, yT);
          if (dist[xS][yS] != -1) {
            int cu = dist[xS][yS];
            if (cu <= largo) {
              string curS = "";
              pair<int, int> po = pa[xS][yS];
              while (po != final) {
                curS += mat[po.first][po.second];
                po = pa[po.first][po.second];
              }
              if (cu < largo) {
                largo = cu;
                best = curS;
              } else if (cu == largo && best > curS) {
                best = curS;
              }
            }
          }
          allow[i] = allow[j] = allow[k] = allow[h] = false;
          if (kk < 4) h = 26;
          if (kk < 3) k = 26;
          if (kk < 2) j = 26;
        }
  if (largo >= 1000000000)
    cout << "-1" << endl;
  else {
    cout << best << endl;
  }
  return 0;
}