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 = 60;
const int maxs = 2600;
const int dx[] = {0, 1, -1, 0};
const int dy[] = {1, 0, 0, -1};
char g[maxn][maxn];
int dist[maxn][maxn];
int sx, sy, tx, ty, n, m, typeCnt;
set<pair<int, int> > ord[maxs];
string ans;
int ansLen;
void work(int stat) {
  queue<int> q;
  q.push(tx);
  q.push(ty);
  memset(dist, -1, sizeof dist);
  dist[tx][ty] = 0;
  for (; !q.empty();) {
    int x = q.front();
    q.pop();
    int y = q.front();
    q.pop();
    for (int i = 0; i < 4; ++i) {
      int xx = x + dx[i], yy = y + dy[i];
      if (xx < 0 || xx >= n || yy < 0 || yy >= m) {
        continue;
      }
      if (g[xx][yy] == 'S' || (stat & (1 << (g[xx][yy] - 'a')))) {
        if (dist[xx][yy] == -1) {
          dist[xx][yy] = dist[x][y] + 1;
          q.push(xx);
          q.push(yy);
        }
      }
    }
  }
  if (dist[sx][sy] == -1 || dist[sx][sy] > ansLen) {
    return;
  }
  string path = "";
  int len = dist[sx][sy];
  for (int i = 0; i <= len; ++i) {
    ord[i].clear();
  }
  ord[len].insert(make_pair(sx, sy));
  for (int i = len; i > 0; --i) {
    char best = 'z' + 1;
    for (set<pair<int, int> >::iterator it = ord[i].begin(); it != ord[i].end();
         ++it) {
      for (int k = 0; k < 4; ++k) {
        int x = it->first + dx[k];
        int y = it->second + dy[k];
        if (x >= 0 && x < n && y >= 0 && y < m && dist[x][y] + 1 == i) {
          if (g[x][y] < best) {
            best = g[x][y];
            ord[i - 1].clear();
          }
          if (g[x][y] == best) {
            ord[i - 1].insert(make_pair(x, y));
          }
        }
      }
    }
    path += best;
  }
  if (len < ansLen || len == ansLen && path < ans) {
    ansLen = len;
    ans = path;
  }
}
void dfs(int i, int j, int stat) {
  if (i == 26) {
    if (j > 0) {
      work(stat);
    }
    return;
  }
  if (j < typeCnt) {
    dfs(i + 1, j + 1, stat | (1 << i));
  }
  dfs(i + 1, j, stat);
}
int main() {
  cin >> n >> m >> typeCnt;
  for (int i = 0; i < n; ++i) {
    cin >> g[i];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (g[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (g[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  ansLen = 1000000000;
  ans = "";
  dfs(0, 0, 0);
  if (ansLen > 1000000) {
    cout << "-1\n";
  } else {
    ans = ans.substr(0, ans.size() - 1);
    cout << ans << endl;
  }
}
