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 maxn = 50 + 5;
int R, C, K;
char g[maxn][maxn];
int sx, sy, tx, ty;
bool flag[300];
int mindist;
string ans;
const int dx[] = {0, 1, -1, 0};
const int dy[] = {1, 0, 0, -1};
bool mark[maxn][maxn];
int dist[maxn][maxn];
string board[maxn][maxn];
set<pair<int, int> > grid[maxn * maxn];
void calc() {
  memset(dist, -1, sizeof dist);
  memset(mark, false, sizeof mark);
  queue<pair<int, int> > q;
  dist[tx][ty] = 0;
  board[tx][ty] = "";
  q.push(make_pair(tx, ty));
  while (!q.empty()) {
    int x = q.front().first;
    int y = q.front().second;
    q.pop();
    for (int i = 0; i < 4; ++i) {
      int xx = x + dx[i], yy = y + dy[i];
      if (xx >= 0 && xx < R && yy >= 0 && yy < C &&
          (flag[g[xx][yy]] || g[xx][yy] == 'S')) {
        if (dist[xx][yy] == -1) {
          dist[xx][yy] = dist[x][y] + 1;
          q.push(make_pair(xx, yy));
        }
      }
    }
  }
  if (dist[sx][sy] == -1) return;
  string ret = "";
  int curd = dist[sx][sy];
  grid[curd].clear();
  grid[curd].insert(make_pair(sx, sy));
  while (curd > 0) {
    char best = 'z' + 1;
    for (set<pair<int, int> >::iterator it = grid[curd].begin();
         it != grid[curd].end(); ++it) {
      for (int i = 0; i < 4; ++i) {
        int xx = it->first + dx[i], yy = it->second + dy[i];
        if (xx >= 0 && xx < R && yy >= 0 && yy < C &&
            dist[xx][yy] + 1 == curd) {
          if (g[xx][yy] < best) {
            best = g[xx][yy];
            grid[curd - 1].clear();
            grid[curd - 1].insert(make_pair(xx, yy));
          } else if (best == g[xx][yy]) {
            grid[curd - 1].insert(make_pair(xx, yy));
          }
        }
      }
    }
    ret += best;
    curd--;
  }
  ret = ret.substr(0, (int)ret.size() - 1);
  if (mindist == -1 || dist[sx][sy] < mindist ||
      dist[sx][sy] == mindist && ret < ans) {
    ans = ret;
    mindist = dist[sx][sy];
  }
}
void dfs(int dep, int last) {
  if (last == 26) {
    if (dep == 0) {
      calc();
    }
    return;
  }
  dfs(dep, last + 1);
  if (dep > 0) {
    flag['a' + last] = 1;
    dfs(dep - 1, last + 1);
    flag['a' + last] = 0;
  }
}
int main() {
  scanf("%d%d%d", &R, &C, &K);
  for (int i = 0; i < R; ++i) {
    scanf("%s", g[i]);
  }
  for (int i = 0; i < R; ++i) {
    for (int j = 0; j < C; ++j) {
      if (g[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (g[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  if (abs(sx - tx) + abs(tx - ty) == 1) {
    puts("");
    return 0;
  }
  ans = "";
  mindist = -1;
  memset(flag, 0, sizeof flag);
  dfs(K, 0);
  if (mindist == -1) {
    printf("%d\n", -1);
  } else {
    puts(ans.c_str());
  }
  return 0;
}
