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;
char tbl[60][60];
int n, m, k, sx, sy, gx, gy;
void input() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", tbl[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (tbl[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (tbl[i][j] == 'T') {
        gx = i;
        gy = j;
      }
}
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
bool ok[130];
int dis[60][60], vis[60][60], fx[60][60], fy[60][60];
inline bool valid(int x, int y) {
  return x >= 0 && x < n && y >= 0 && y < m && ok[(int)tbl[x][y]];
}
bool cmp(int x1, int y1, int x2, int y2) {
  if (tbl[x1][y1] != tbl[x2][y2]) return tbl[x1][y1] < tbl[x2][y2];
  if (x1 == y1 && x2 == y2) return 0;
  return cmp(fx[x1][y1], fy[x1][y1], fx[x2][y2], fy[x2][y2]);
}
string go(int a, int b, int c, int d) {
  static int tmt = 0;
  tmt++;
  ok[a] = ok[b] = ok[c] = ok[d] = 1;
  queue<int> q;
  vis[gx][gy] = tmt;
  dis[gx][gy] = 0;
  q.push(gx);
  q.push(gy);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    int y = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int nx = x + dx[i];
      int ny = y + dy[i];
      if (!valid(nx, ny)) continue;
      if (vis[nx][ny] != tmt) {
        vis[nx][ny] = tmt;
        dis[nx][ny] = dis[x][y] + 1;
        fx[nx][ny] = x;
        fy[nx][ny] = y;
        q.push(nx);
        q.push(ny);
      } else if (dis[nx][ny] == dis[x][y] + 1) {
        if (cmp(x, y, fx[nx][ny], fy[nx][ny])) {
          fx[nx][ny] = x;
          fy[nx][ny] = y;
        }
      }
    }
  }
  ok[a] = ok[b] = ok[c] = ok[d] = 0;
  if (vis[sx][sy] != tmt)
    return "!";
  else {
    string ret;
    int x = fx[sx][sy], y = fy[sx][sy], xx, yy;
    while (x != gx || y != gy) {
      ret += tbl[x][y];
      xx = fx[x][y];
      yy = fy[x][y];
      x = xx;
      y = yy;
    }
    return ret;
  }
}
int len = 10000;
string ans;
void chk(int a, int b, int c, int d) {
  string now = go(a, b, c, d);
  if (now == "!") return;
  if ((int)now.size() < len || ((int)now.size() == len && now < ans)) {
    ans = now;
    len = now.size();
  }
}
void solve() {
  for (int a = 'a'; a <= 'z'; a++)
    if (k == 1)
      chk(a, 0, 0, 0);
    else
      for (int b = a + 1; b <= 'z'; b++)
        if (k == 2)
          chk(a, b, 0, 0);
        else
          for (int c = b + 1; c <= 'z'; c++)
            if (k == 3)
              chk(a, b, c, 0);
            else
              for (int d = c + 1; d <= 'z'; d++) chk(a, b, c, d);
  if (len == 10000)
    puts("-1");
  else
    printf("%s\n", ans.c_str());
}
int main() {
  ok[(int)'S'] = ok[(int)'T'] = 1;
  input();
  solve();
  return 0;
}
