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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, m, k;
string a[57];
bool used[27];
string ans = "Z";
int d[57][57];
bool b[57][57];
vector<pair<int, int> > v[2];
bool ok(int x, int y) { return x >= 0 && y >= 0 && x < n && y < m; }
string run() {
  int x, y;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j] == 'T') x = i, y = j;
  memset(d, 0, sizeof(d));
  queue<pair<int, int> > q;
  q.push(make_pair(x, y));
  d[x][y] = 1;
  while (!q.empty()) {
    x = q.front().first;
    y = q.front().second;
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (!ok(x + dx[i], y + dy[i])) continue;
      if (d[x + dx[i]][y + dy[i]] != 0) continue;
      char c = a[x + dx[i]][y + dy[i]];
      if (!used[c - 'a'] && c != 'S' && c != 'T') continue;
      d[x + dx[i]][y + dy[i]] = d[x][y] + 1;
      q.push(make_pair(x + dx[i], y + dy[i]));
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j] == 'S') x = i, y = j;
  if (d[x][y] == 0) return "Z";
  memset(b, false, sizeof(b));
  int cur = 0, prev = 1;
  v[cur].push_back(make_pair(x, y));
  b[x][y] = true;
  string res = "";
  while (!v[cur].empty()) {
    cur = 1 - cur;
    prev = 1 - prev;
    v[cur].clear();
    char c = 'z' + 1;
    for (int i = 0; i < (int)v[prev].size(); i++) {
      x = v[prev][i].first;
      y = v[prev][i].second;
      for (int j = 0; j < 4; j++) {
        if (!ok(x + dx[j], y + dy[j])) continue;
        if (d[x + dx[j]][y + dy[j]] + 1 != d[x][y]) continue;
        if (d[x + dx[j]][y + dy[j]] == 0) continue;
        c = min(c, a[x + dx[j]][y + dy[j]]);
      }
    }
    for (int i = 0; i < (int)v[prev].size(); i++) {
      x = v[prev][i].first;
      y = v[prev][i].second;
      for (int j = 0; j < 4; j++) {
        if (!ok(x + dx[j], y + dy[j])) continue;
        if (d[x + dx[j]][y + dy[j]] + 1 != d[x][y]) continue;
        if (d[x + dx[j]][y + dy[j]] == 0) continue;
        if (a[x + dx[j]][y + dy[j]] != c) continue;
        if (b[x + dx[j]][y + dy[j]]) continue;
        b[x + dx[j]][y + dy[j]] = true;
        v[cur].push_back(make_pair(x + dx[j], y + dy[j]));
      }
    }
    if (c >= 'a' && c <= 'z') res += c;
  }
  return res;
}
void dfs(int x = 0, int K = 0) {
  if (K == k) {
    string res = run();
    if (res != "Z" && (ans == "Z" || res.size() < ans.size() ||
                       (res.size() == ans.size() && res < ans)))
      ans = res;
    return;
  }
  for (int i = x; i < 26; i++)
    used[i] = true, dfs(i + 1, K + 1), used[i] = false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  dfs();
  if (ans == "Z") ans = "-1";
  cout << ans;
  return 0;
}