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;
int k;
unordered_map<char, int> used;
const int N = 55;
int inf = 1e9, dp[N][N], n, m, sx, sy;
char s[N][N];
int tx, ty, best = inf, pos[N][N];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
string BEST = "-1";
void solve() {
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j] = inf;
      pos[i][j] = 0;
    }
  }
  dp[tx][ty] = 0;
  queue<pair<int, int> > q;
  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 nx = x + dx[i];
      int ny = y + dy[i];
      if (used[s[nx][ny]] && dp[nx][ny] == inf) {
        dp[nx][ny] = dp[x][y] + 1;
        q.push(make_pair(nx, ny));
      }
    }
  }
  if (dp[sx][sy] == inf || dp[sx][sy] > best) {
    return;
  }
  best = dp[sx][sy];
  vector<pair<int, int> > v;
  v.push_back(make_pair(sx, sy));
  pos[sx][sy] = 1;
  string ans;
  while (1) {
    vector<pair<int, int> > nv;
    char mn = 'z' + 1;
    for (auto i : v) {
      int x = i.first;
      int y = i.second;
      for (int j = 0; j < 4; j++) {
        int nx = x + dx[j];
        int ny = y + dy[j];
        if (dp[x][y] - 1 == dp[nx][ny] && used[s[nx][ny]]) {
          if (mn > s[nx][ny] && !pos[nx][ny]) {
            nv.clear();
            nv.push_back(make_pair(nx, ny));
            mn = s[nx][ny];
            pos[nx][ny] = 1;
          } else if (mn == s[nx][ny] && !pos[nx][ny]) {
            nv.push_back(make_pair(nx, ny));
            pos[nx][ny] = 1;
          }
        }
      }
    }
    if (mn == 'T') {
      if (BEST == "-1" || BEST.size() > ans.size() || BEST > ans) {
        BEST = ans;
      }
      return;
    }
    ans += mn;
    v = nv;
  }
}
void rec(int x, int cnt) {
  if (x == 26) {
    solve();
    return;
  }
  used[char(x + 'a')] = 0;
  rec(x + 1, cnt);
  if (cnt == k) return;
  used[char(x + 'a')] = 1;
  rec(x + 1, cnt + 1);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i] + 1);
    for (int j = 1; j <= m; j++) {
      if (s[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (s[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  used['S'] = used['T'] = 1;
  rec(0, 0);
  cout << BEST;
  return 0;
}
