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 a1[4] = {0, 0, 1, -1}, a2[4] = {1, -1, 0, 0};
struct ww {
  int x, y;
  inline bool operator==(const ww &A) { return x == A.x && y == A.y; }
} S, T, g[60 * 60];
struct w {
  int a[15];
  int len;
  inline bool operator<(const w &A) {
    if (len < A.len) return 1;
    if (len > A.len) return 0;
    int i;
    for (i = len / 15; i >= 0; i--) {
      if (a[i] < A.a[i]) return 1;
      if (a[i] > A.a[i]) return 0;
    }
    return 1;
  }
} gg[60][60];
struct Ans {
  int len;
  char p[100];
  inline bool operator<(const Ans &A) {
    if (len < A.len) return 1;
    if (len > A.len) return 0;
    int i;
    for (i = 1; i <= len; i++) {
      if (p[i] < A.p[i]) return 1;
      if (p[i] > A.p[i]) return 0;
    }
    return 1;
  }
  inline void print() {
    if (len == 100)
      printf("-1\n");
    else {
      int i;
      for (i = 1; i <= len; i++) printf("%c", p[i]);
      printf("\n");
    }
  }
} an;
int i, j, k, n, m, K;
int id[60], a[60];
char p[60][60], re[100];
inline void into(w &A, int B) {
  int a1 = A.len / 15, a2 = A.len % 15;
  A.len++;
  A.a[a1] |= B << (2 * a2);
}
inline void work() {
  int i, j, l, r;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) gg[i][j].len = 1000;
  g[l = r = 1] = T;
  gg[T.x][T.y].len = 0;
  for (i = 0; i <= 14; i++) gg[T.x][T.y].a[i] = 0;
  for (; l <= r; l++) {
    ww A = g[l];
    if (A == S) break;
    for (i = 0; i <= 3; i++) {
      int x = A.x + a1[i], y = A.y + a2[i];
      ww asd = {x, y};
      if (x && y && x <= n && y <= m) {
        int B = id[p[x][y] - 'a' + 1];
        if (!(asd == S) && !B) continue;
        w C = gg[A.x][A.y];
        if (!(asd == S)) into(C, B - 1);
        if (C < gg[x][y]) {
          if (gg[x][y].len == 1000) g[++r] = asd;
          gg[x][y] = C;
        }
      }
    }
  }
  if (l > r) return;
  w res = gg[S.x][S.y];
  if (res.len > an.len) return;
  Ans ans = {};
  for (i = res.len - 1; i >= 0; i--) {
    int A = i / 15, B = i % 15;
    int C = (res.a[A] >> (2 * B)) & 3;
    ans.p[++ans.len] = 'a' + a[C + 1] - 1;
  }
  if (ans < an) an = ans;
}
void dfs(int x, int y) {
  if (x > K) {
    work();
    return;
  }
  int i;
  for (i = y + 1; i <= 26; i++) {
    id[i] = x;
    a[x] = i;
    dfs(x + 1, i);
    id[i] = 0;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= n; i++) scanf("%s", p[i] + 1);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      if (p[i][j] == 'S') S = (ww){i, j};
      if (p[i][j] == 'T') T = (ww){i, j};
    }
  an.len = 100;
  dfs(1, 0);
  an.print();
  return 0;
}
