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;
int n, m, k;
char mp[60][60];
bool can[60][60];
int dist[60][60];
pair<int, int> src, trg;
queue<pair<int, int> > q;
vector<pair<int, int> > pos[300];
int dir[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
int best = 1000000000;
char ans[1000000];
char tans[1000000];
pair<int, int> posi[2][1000000];
int sz[2];
void f(int rem, char st) {
  if (rem == 0) {
    memset(dist, -1, sizeof dist);
    while (!q.empty()) q.pop();
    q.push(trg);
    dist[trg.first][trg.second] = 0;
    while (!q.empty()) {
      pair<int, int> tmp = q.front();
      q.pop();
      if (tmp == src) break;
      for (int i = 0; i < 4; i++) {
        int nr = tmp.first + dir[i][0];
        int nc = tmp.second + dir[i][1];
        if (nr >= 0 && nc >= 0 && nr < n && nc < m && can[nr][nc] &&
            dist[nr][nc] == -1) {
          dist[nr][nc] = dist[tmp.first][tmp.second] + 1;
          q.push(pair<int, int>(nr, nc));
        }
      }
    }
    int len = dist[src.first][src.second];
    if (len != -1 && len <= best) {
      int cur = 0, nx = 1;
      posi[0][0] = src;
      sz[0] = 1;
      for (int i = len - 1; i >= 0; i--) {
        sz[nx] = 0;
        char mn = 'z' + 1;
        for (int j = 0; j < sz[cur]; j++) {
          for (int k = 0; k < 4; k++) {
            int nr = posi[cur][j].first + dir[k][0];
            int nc = posi[cur][j].second + dir[k][1];
            if (nr >= 0 && nc >= 0 && nr < n && nc < m && dist[nr][nc] == i) {
              mn = min(mn, mp[nr][nc]);
            }
          }
        }
        for (int j = 0; j < sz[cur]; j++) {
          for (int k = 0; k < 4; k++) {
            int nr = posi[cur][j].first + dir[k][0];
            int nc = posi[cur][j].second + dir[k][1];
            if (nr >= 0 && nc >= 0 && nr < n && nc < m && dist[nr][nc] == i &&
                mp[nr][nc] == mn) {
              posi[nx][sz[nx]++] = pair<int, int>(nr, nc);
            }
          }
        }
        sort(posi[nx], posi[nx] + sz[nx]);
        int ptr = 0;
        for (int j = 0; j < sz[nx]; j++) {
          if (j == 0 || posi[nx][j] != posi[nx][j - 1])
            posi[nx][ptr++] = posi[nx][j];
        }
        sz[nx] = ptr;
        tans[len - i - 1] = mn;
        cur = 1 - cur, nx = 1 - nx;
      }
      bool good = true;
      if (best == len) {
        for (int i = 0; i < len; i++)
          if (tans[i] != ans[i]) {
            if (tans[i] < ans[i])
              break;
            else
              good = false;
          }
      }
      if (good) {
        for (int i = 0; i < len; i++) ans[i] = tans[i];
        best = len;
      }
    }
    return;
  }
  for (char i = st; i <= 'z'; i++) {
    for (int j = 0; j < pos[i].size(); j++)
      can[pos[i][j].first][pos[i][j].second] = true;
    f(rem - 1, i + 1);
    for (int j = 0; j < pos[i].size(); j++)
      can[pos[i][j].first][pos[i][j].second] = false;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mp[i][j];
      if (mp[i][j] == 'S')
        src = pair<int, int>(i, j);
      else if (mp[i][j] == 'T')
        trg = pair<int, int>(i, j);
      else
        pos[mp[i][j]].push_back(pair<int, int>(i, j));
    }
  }
  can[src.first][src.second] = true;
  f(k, 'a');
  if (best == 1000000000)
    cout << -1 << endl;
  else {
    ans[best - 1] = 0;
    puts(ans);
  }
  return 0;
}