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 MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN(55);
struct nnode {
  int x, y, d;
  string s;
  nnode(){};
  nnode(int x, int y, int d, string s) : x(x), y(y), d(d), s(s){};
  bool operator<(const nnode &other) const {
    if (d != other.d) return d > other.d;
    return s > other.s;
  }
};
char a[MAXN][MAXN];
int n, m, k, path[MAXN][MAXN], available[26], ans;
int d[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
pair<int, int> q[MAXN * MAXN];
string ansPath, p[MAXN][MAXN];
void SSSP() {
  int fr = 0, bk = 0;
  pair<int, int> s, t;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      path[i][j] = INF;
      p[i][j] = "";
      if (a[i][j] == 'S') {
        s = pair<int, int>(i, j);
      } else if (a[i][j] == 'T') {
        t = pair<int, int>(i, j);
        q[bk++] = t;
        path[t.first][t.second] = 0;
      }
    }
  vector<pair<int, int> > b;
  while (fr != bk) {
    int x = q[fr].first, y = q[fr].second;
    fr++;
    b.push_back(pair<int, int>(x, y));
    if (a[x][y] == 'S') break;
    for (int i = 0; i < 4; i++) {
      int tx = x + d[i][0];
      int ty = y + d[i][1];
      if (tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
      if (a[tx][ty] == 'T') continue;
      if (a[tx][ty] != 'S' && !available[a[tx][ty] - 'a']) continue;
      if (path[x][y] + 1 < path[tx][ty]) {
        q[bk++] = pair<int, int>(tx, ty);
        path[tx][ty] = path[x][y] + 1;
      }
    }
  }
  if (path[s.first][s.second] == INF) return;
  p[s.first][s.second] = "S";
  for (int i = (int)b.size() - 1; i >= 0; i--) {
    int x = b[i].first, y = b[i].second;
    if (p[s.first][s.second] == "") continue;
    for (int j = 0; j < 4; j++) {
      int tx = x + d[j][0];
      int ty = y + d[j][1];
      if (tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
      if (path[x][y] - 1 == path[tx][ty]) {
        if (p[tx][ty] == "")
          p[tx][ty] = p[x][y] + a[tx][ty];
        else
          p[tx][ty] = min(p[tx][ty], p[x][y] + a[tx][ty]);
      }
    }
  }
  if (path[s.first][s.second] < ans ||
      (path[s.first][s.second] == ans && p[t.first][t.second] < ansPath)) {
    ans = path[s.first][s.second];
    ansPath = p[t.first][t.second];
  }
}
void combination(int now, int i) {
  if (now == k) {
    SSSP();
    return;
  }
  while (i < 26) {
    available[i] = 1;
    combination(now + 1, i + 1);
    available[i] = 0;
    i++;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", a[i]);
  ans = INF;
  ansPath = "";
  combination(0, -1);
  if (ans == INF)
    printf("-1");
  else
    cout << ansPath.substr(1, ansPath.size() - 2);
}
