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;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int x = 0, p = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * p;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * x % mod;
    x = x * x % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
  return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
  if (B == 0) return A;
  return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
  return A * B / gcd(A, B);
}
};  // namespace Mymath
const int Maxn = 55;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
char c[Maxn][Maxn];
int n, m, k, sx, sy, ex, ey;
vector<vector<int> > V;
vector<int> now;
void dfs(int lvl) {
  if (lvl == k) {
    V.push_back(now);
    return;
  }
  int lst = (now.empty()) ? 0 : now.back() + 1;
  for (int i = lst; i < 26; i++) {
    now.push_back(i);
    dfs(lvl + 1);
    now.pop_back();
  }
}
int ds[Maxn][Maxn], dt[Maxn][Maxn];
bool G[27];
int qx[Maxn * Maxn], qy[Maxn * Maxn], eq, kq;
void bfs(int x, int y, int dist[Maxn][Maxn]) {
  for (int i = 0; i < Maxn; i++) {
    for (int j = 0; j < Maxn; j++) {
      dist[i][j] = -1;
    }
  }
  dist[x][y] = 0;
  kq = eq = 0;
  qx[eq] = x;
  qy[eq++] = y;
  while (kq < eq) {
    int X = qx[kq], Y = qy[kq++];
    for (int i = 0; i < 4; i++) {
      int nx = X + dx[i], ny = Y + dy[i];
      if (nx >= 0 && ny >= 0 && nx < n && ny < m &&
          (c[nx][ny] == 'S' || c[nx][ny] == 'T' || G[c[nx][ny] - 'a'])) {
        if (dist[nx][ny] == -1) {
          dist[nx][ny] = dist[X][Y] + 1;
          qx[eq] = nx;
          qy[eq++] = ny;
        }
      }
    }
  }
}
int anslen = 1e9;
string ans;
bool vis[Maxn][Maxn];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", c[i]);
    for (int j = 0; j < m; j++) {
      if (c[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (c[i][j] == 'T') {
        ex = i;
        ey = j;
      }
    }
  }
  dfs(0);
  for (int i = 0; i < V.size(); i++) {
    memset(G, 0, sizeof(G));
    memset(vis, 0, sizeof(vis));
    for (int j = 0; j < V[i].size(); j++) {
      G[V[i][j]] = 1;
    }
    bfs(sx, sy, ds);
    bfs(ex, ey, dt);
    if (ds[ex][ey] == -1) continue;
    string cur;
    int dst = ds[ex][ey];
    vector<pair<int, int> > Now, Nxt;
    char cc = 'z' + 1;
    Now.push_back(make_pair(sx, sy));
    for (int j = 1; j < dst; j++) {
      cc = 'z' + 1;
      for (int t = 0; t < Now.size(); t++) {
        int x = Now[t].first, y = Now[t].second;
        for (int d = 0; d < 4; d++) {
          int nx = x + dx[d], ny = y + dy[d];
          if (nx >= 0 && ny >= 0 && nx < n && ny < m && !vis[nx][ny] &&
              j + dt[nx][ny] == dst) {
            vis[nx][ny] = 1;
            char now = c[nx][ny];
            if (now < cc) {
              Nxt.clear();
              cc = now;
            }
            if (now == cc) {
              Nxt.push_back(make_pair(nx, ny));
            }
          }
        }
      }
      cur += cc;
      Now = Nxt;
      Nxt.clear();
    }
    if (anslen > dst) {
      anslen = dst;
      ans = cur;
    } else if (anslen == dst) {
      if (cur < ans) ans = cur;
    }
  }
  if (anslen == 1e9) {
    printf("-1\n");
    return 0;
  }
  printf("%s\n", ans.c_str());
  return 0;
}