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;
template <class T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <class T>
void pvp(T a, T b) {
  for (T i = a; i != b; ++i)
    cout << "(" << i->first << ", " << i->second << ") ";
  cout << endl;
}
template <class T>
void chmin(T &t, T f) {
  if (t > f) t = f;
}
template <class T>
void chmax(T &t, T f) {
  if (t < f) t = f;
}
int in_c() {
  int c;
  for (; (c = getchar()) <= ' ';) {
    if (!~c) throw ~0;
  }
  return c;
}
int in() {
  int x = 0, c;
  for (; (unsigned)((c = getchar()) - '0') >= 10;) {
    if (c == '-') return -in();
    if (!~c) throw ~0;
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while ((unsigned)((c = getchar()) - '0') < 10);
  return x;
}
long long In() {
  long long x = 0, c;
  for (; (unsigned)((c = getchar()) - '0') >= 10;) {
    if (c == '-') return -In();
    if (!~c) throw ~0;
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while ((unsigned)((c = getchar()) - '0') < 10);
  return x;
}
int dx[] = {
    1,
    0,
    -1,
    0,
};
int dy[] = {
    0,
    1,
    0,
    -1,
};
const int L = 26;
int M, N, K;
char S[60][60];
int SX, SY, TX, TY;
int sho;
string ans;
int safe[60];
int csLen, cs[60];
int que[100010], *qb, *qe;
int d[60][60];
bool vis[60][60];
string calcIt() {
  csLen = 0;
  for (int l = 0; l < L; ++l)
    if (safe[l]) cs[csLen++] = l;
  assert(csLen == K);
  int k;
  int x, y;
  int h, xx, yy;
  int *qq, *qnxt;
  qb = qe = que;
  memset(vis, 0, sizeof(vis));
  vis[SX][SY] = 1;
  *qe++ = SX;
  *qe++ = SY;
  string ret = "";
  for (;;) {
    for (k = 0; k < K; ++k) {
      bool ok = 0;
      qnxt = qe;
      for (qq = qb; qq != qe;) {
        x = *qq++;
        y = *qq++;
        for (h = 0; h < 4; ++h) {
          xx = x + dx[h];
          yy = y + dy[h];
          if (0 <= xx && xx < M && 0 <= yy && yy < N) {
            if (S[xx][yy] == 'T') {
              return ret;
            }
            if (!vis[xx][yy] && d[x][y] == d[xx][yy] + 1) {
              if (S[xx][yy] - 'a' == cs[k]) {
                ok = 1;
                vis[xx][yy] = 1;
                *qnxt++ = xx;
                *qnxt++ = yy;
              }
            }
          }
        }
      }
      if (ok) {
        ret += (char)('a' + cs[k]);
        qb = qe;
        qe = qnxt;
        break;
      }
    }
    assert(k < K);
  }
}
void solve(int flg) {
  int x, y;
  int h, xx, yy;
  qb = qe = que;
  memset(d, ~0, sizeof(d));
  d[TX][TY] = 0;
  *qe++ = TX;
  *qe++ = TY;
  for (; qb != qe;) {
    x = *qb++;
    y = *qb++;
    for (h = 0; h < 4; ++h) {
      xx = x + dx[h];
      yy = y + dy[h];
      if (0 <= xx && xx < M && 0 <= yy && yy < N) {
        if (S[xx][yy] == 'S') {
          d[xx][yy] = d[x][y] + 1;
          goto found;
        }
        if (safe[S[xx][yy] - 'a']) {
          if (!~d[xx][yy]) {
            d[xx][yy] = d[x][y] + 1;
            *qe++ = xx;
            *qe++ = yy;
          }
        }
      }
    }
  }
  return;
found:;
  if (flg == 0) {
    chmin(sho, d[SX][SY]);
    return;
  } else {
    if (sho != d[SX][SY]) return;
  }
  string s = calcIt();
  if (ans == "~" || ans.size() > s.size() ||
      ans.size() == s.size() && ans > s) {
    ans = s;
  }
}
void dfs(int k, int a, int flg) {
  if (k == K) {
    solve(flg);
    return;
  }
  for (; a < L; ++a) {
    safe[a] = 1;
    dfs(k + 1, a + 1, flg);
    safe[a] = 0;
  }
}
int main() {
  int x, y;
  for (; ~scanf("%d%d%d", &M, &N, &K);) {
    for (x = 0; x < M; ++x) {
      scanf("%s", S[x]);
    }
    for (x = 0; x < M; ++x)
      for (y = 0; y < N; ++y) {
        if (S[x][y] == 'S') SX = x, SY = y;
        if (S[x][y] == 'T') TX = x, TY = y;
      }
    sho = M * N;
    ans = "~";
    dfs(0, 0, 0);
    dfs(0, 0, 1);
    if (ans == "~") ans = "-1";
    if (ans[0] == 'S') ans = ans.substr(1);
    cout << ans << endl;
  }
  return 0;
}