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 INF = 1e9, mf[4] = {-1, 1, 0, 0}, mc[4] = {0, 0, -1, 1};
int n, m, k, msk, fi, ci, ff, cf, dp[55][55], mk[55][55], res = INF;
char a[55][55];
string ans = "{";
struct data {
  int f, c;
  data(int f, int c) : f(f), c(c) {}
};
void bfs() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) dp[i][j] = INF;
  queue<data> q;
  q.push(data(ff, cf));
  dp[ff][cf] = 0;
  while (!q.empty()) {
    int f = q.front().f;
    int c = q.front().c;
    q.pop();
    if (a[f][c] == 'S') break;
    for (int i = 0; i < 4; i++) {
      int nf = f + mf[i];
      int nc = c + mc[i];
      if (nf < 1 || nf > n || nc < 1 || nc > m || a[nf][nc] == 'T') continue;
      int id = a[nf][nc] - 'a';
      if (a[nf][nc] == 'S' || (msk & 1 << id)) {
        if (dp[nf][nc] == INF) {
          dp[nf][nc] = dp[f][c] + 1;
          q.push(data(nf, nc));
        }
      }
    }
  }
}
string relax() {
  memset(mk, 0, sizeof(mk));
  queue<data> q;
  vector<data> v;
  q.push(data(fi, ci));
  mk[fi][ci] = 1;
  string w = "";
  while (1) {
    while (!q.empty()) {
      v.push_back(q.front());
      q.pop();
    }
    int finish = 0;
    char mn = '{';
    for (auto o : v) {
      int f = o.f;
      int c = o.c;
      for (int i = 0; i < 4; i++) {
        int nf = f + mf[i];
        int nc = c + mc[i];
        if (nf < 1 || nf > n || nc < 1 || nc > m || a[nf][nc] == 'S' ||
            mk[nf][nc])
          continue;
        if (dp[nf][nc] == INF) continue;
        int id = a[nf][nc] - 'a';
        if (a[nf][nc] == 'T') {
          finish = 1;
          break;
        }
        if ((msk & 1 << id) && dp[nf][nc] + 1 == dp[f][c])
          mn = min(mn, a[nf][nc]);
      }
      if (finish) break;
    }
    if (finish) break;
    int x = 1e9, y = -1e9;
    for (auto o : v) {
      int f = o.f;
      int c = o.c;
      for (int i = 0; i < 4; i++) {
        int nf = f + mf[i];
        int nc = c + mc[i];
        if (nf < 1 || nf > n || nc < 1 || nc > m || a[nf][nc] == 'S' ||
            mk[nf][nc])
          continue;
        if (dp[nf][nc] == INF) continue;
        int id = a[nf][nc] - 'a';
        if ((msk & 1 << id) && dp[nf][nc] + 1 == dp[f][c] && a[nf][nc] == mn) {
          if (!mk[nf][nc]) {
            x = min(x, dp[nf][nc]);
            y = max(y, dp[nf][nc]);
            q.push(data(nf, nc));
          }
          mk[nf][nc] = 1;
        }
      }
    }
    w.push_back(mn);
    v.clear();
  }
  return w;
}
void rec(int p, int t) {
  if (t == k) {
    bfs();
    if (dp[fi][ci] == INF) return;
    if (dp[fi][ci] < res) {
      res = dp[fi][ci];
      ans = relax();
    } else if (dp[fi][ci] == res)
      ans = min(ans, relax());
    return;
  }
  for (int i = p; i < 26; i++) {
    msk |= 1 << i;
    rec(i + 1, t + 1);
    msk ^= 1 << i;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%s", a[i] + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == 'S') fi = i, ci = j;
      if (a[i][j] == 'T') ff = i, cf = j;
    }
  }
  rec(0, 0);
  cout << (ans == "{" ? "-1" : ans) << '\n';
  return 0;
}
