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 n, m, k, can[256], fx, fy, sx, sy;
int d[] = {0, 1, 1, 0, -1, 0, 0, -1};
int a[70][70], ans = (1 << 30);
char Map[70][70];
bool u[70][70];
queue<int> qx, qy, Qx[2], Qy[2];
string strans, str;
bool check(int x, int y) { return x >= 0 && x < n && y >= 0 && y < m; }
void path(int steps) {
  str = "";
  steps--;
  int x, y, dx, dy, mn;
  char want;
  Qx[steps & 1].push(sx);
  Qy[steps & 1].push(sy);
  for (; steps >= 0; steps--) {
    queue<int> &qX = Qx[steps & 1], &qY = Qy[steps & 1],
               &nx = Qx[(steps + 1) & 1], &ny = Qy[(steps + 1) & 1];
    want = 'z' + 1;
    memset(u, 0, sizeof u);
    while (!qX.empty()) {
      x = qX.front();
      y = qY.front();
      qX.pop();
      qY.pop();
      for (int i = 0; i < 8; i += 2) {
        dx = x + d[i];
        dy = y + d[i + 1];
        if (!check(dx, dy)) continue;
        if (a[dx][dy] != a[x][y] - 1) continue;
        if (Map[dx][dy] < want) {
          while (!nx.empty()) nx.pop(), ny.pop();
          want = Map[dx][dy];
        }
        if (Map[dx][dy] == want && !u[dx][dy])
          nx.push(dx), ny.push(dy), u[dx][dy] = 1;
      }
    }
    if (want == 'T') break;
    str.push_back(want);
  }
}
void solve() {
  int x, y, dx, dy;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      a[i][j] = (1 << 30);
    }
  }
  qx.push(fx);
  qy.push(fy);
  a[fx][fy] = 0;
  while (!qx.empty()) {
    x = qx.front();
    y = qy.front();
    qx.pop();
    qy.pop();
    for (int i = 0; i < 8; i += 2) {
      dx = x + d[i];
      dy = y + d[i + 1];
      if (check(dx, dy) && a[dx][dy] > a[x][y] + 1 && can[Map[dx][dy]]) {
        a[dx][dy] = a[x][y] + 1;
        qx.push(dx);
        qy.push(dy);
      }
    }
  }
  if (a[sx][sy] == (1 << 30)) return;
  if (a[sx][sy] < ans) {
    ans = a[sx][sy];
    path(a[sx][sy]);
    strans = str;
  } else if (a[sx][sy] == ans) {
    path(a[sx][sy]);
    if (str < strans) strans = str;
  }
}
void rec(int d, int i = 0) {
  if (d == k) {
    solve();
    return;
  }
  for (; i < 27; i++) {
    can['a' + i] = 1;
    rec(d + 1, i + 1);
    can['a' + i] = 0;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", Map[i]);
    for (int j = 0; j < m; j++) {
      if (Map[i][j] == 'S') sx = i, sy = j;
      if (Map[i][j] == 'T') fx = i, fy = j;
    }
  }
  can['S'] = can['T'] = 1;
  rec(0);
  if (ans < (1 << 30))
    printf("%s\n", strans.c_str());
  else
    printf("-1\n");
  return 0;
}
