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;
string a[52];
int dr[] = {-1, 0, 1, 0};
int dc[] = {0, -1, 0, 1};
int sr, sc, tr, tc;
string ans(52 * 52, ' ');
bool allowed['z' + 1] = {0};
bool vis[52][52];
int d[52][52];
void bfs() {
  for (char i = 'a'; i <= 'z'; ++i)
    if (allowed[(int)i]) clog << i;
  clog << endl;
  static queue<pair<char, char> > qu;
  memset(d, -1, sizeof(d));
  d[tr][tc] = 0;
  for (qu.push({tr, tc}); qu.size(); qu.pop()) {
    int r = qu.front().first, c = qu.front().second;
    for (int i = -1; ++i < 4;) {
      int tr = r + dr[i], tc = c + dc[i];
      if (tr < 0 or tr >= n or tc < 0 or tc >= m) continue;
      if (d[tr][tc] != -1) continue;
      if (a[tr][tc] == 'S' or allowed[(int)a[tr][tc]]) d[tr][tc] = d[r][c] + 1;
      if (!allowed[(int)a[tr][tc]]) continue;
      qu.push({tr, tc});
    }
  }
  if (d[sr][sc] == -1) return;
  static vector<pair<char, char> > curlay, nxtlay;
  curlay.clear();
  nxtlay.clear();
  curlay.push_back({sr, sc});
  string tans = "";
  memset(vis, 0, sizeof(vis));
  for (int i = -1; ++i < d[sr][sc] - 1;) {
    char minchar = 'z' + 1;
    for (; curlay.size(); curlay.pop_back()) {
      int r = curlay.back().first, c = curlay.back().second;
      for (int i = -1; ++i < 4;) {
        int tr = r + dr[i], tc = c + dc[i];
        if (tr < 0 or tr >= n or tc < 0 or tc >= m) continue;
        if (vis[tr][tc]) continue;
        if (a[tr][tc] != 'S' and a[tr][tc] != 'T' and !allowed[(int)a[tr][tc]])
          continue;
        if (d[tr][tc] != d[r][c] - 1) continue;
        vis[tr][tc] = 1;
        if (a[tr][tc] < minchar) {
          minchar = a[tr][tc];
          nxtlay.clear();
        }
        if (a[tr][tc] == minchar) nxtlay.push_back({tr, tc});
      }
    }
    tans += minchar;
    curlay = nxtlay;
    nxtlay.clear();
  }
  if (((int)ans.size()) < ((int)tans.size())) return;
  if (((int)ans.size()) > ((int)tans.size()) or ans > tans) ans = tans;
}
void process(int i = 'a') {
  if (k == 0) {
    bfs();
    return;
  }
  if (i > 'z') return;
  for (int t = -1; ++t < 2;) {
    k -= t;
    allowed[i] = t;
    process(i + 1);
    allowed[i] = 0;
    k += t;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = -1; ++i < n;) cin >> a[i];
  for (int i = -1; ++i < n;)
    for (int f = -1; ++f < m;) {
      if (a[i][f] == 'S') sr = i, sc = f;
      if (a[i][f] == 'T') tr = i, tc = f;
    }
  process();
  if (((int)ans.size()) > m * n) ans = "-1";
  cout << ans;
  return 0;
}