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 N = 55;
string s[N][N], a[N], ans;
int n, m, k, dis[N][N], x1, z, d = 1e9, x, y, e1, e2;
bool mark[N][N];
vector<pair<int, int> > l;
int X[4] = {0, 1, 0, -1};
int Y[4] = {1, 0, -1, 0};
bool valid(int f1, int f2) {
  if (f1 <= 0 or f2 <= 0 or f1 > n or f2 > m or !mark[f1][f2]) return false;
  return true;
}
void bfs() {
  mark[e1][e2] = 1;
  l.clear();
  l.push_back({x1, z});
  dis[x1][z] = 0;
  for (int i = 0; i < l.size(); i++) {
    x = l[i].first, y = l[i].second;
    for (int j = 0; j < 4; j++) {
      int xnow = X[j] + x, ynow = Y[j] + y;
      if (valid(xnow, ynow)) {
        if (dis[xnow][ynow] == dis[x][y] + 1)
          s[xnow][ynow] = min(s[xnow][ynow], s[x][y] + a[xnow][ynow - 1]);
        else if (dis[xnow][ynow] > dis[x][y]) {
          dis[xnow][ynow] = dis[x][y] + 1;
          s[xnow][ynow] = s[x][y] + a[xnow][ynow - 1];
          l.push_back({xnow, ynow});
        }
      }
    }
  }
  if (dis[e1][e2] == d)
    ans = min(ans, s[e1][e2]);
  else if (dis[e1][e2] < d) {
    ans = s[e1][e2];
    d = dis[e1][e2];
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j] == 'S')
        x1 = i, z = j + 1;
      else if (a[i][j] == 'T')
        e1 = i, e2 = j + 1;
  if (k == 4) {
    for (int i1 = 0; i1 < 26; i1++) {
      for (int i2 = i1 + 1; i2 < 26; i2++) {
        for (int i3 = i2 + 1; i3 < 26; i3++) {
          for (int i4 = i3 + 1; i4 < 26; i4++) {
            for (int i = 1; i <= n; i++)
              for (int j = 1; j <= m; j++) {
                if (a[i][j - 1] - 'a' == i1 or a[i][j - 1] - 'a' == i2 or
                    a[i][j - 1] - 'a' == i3 or a[i][j - 1] - 'a' == i4)
                  mark[i][j] = 1;
                else
                  mark[i][j] = 0;
                dis[i][j] = 1e9;
              }
            bfs();
          }
        }
      }
    }
  }
  if (k == 3) {
    for (int i1 = 0; i1 < 26; i1++) {
      for (int i2 = i1 + 1; i2 < 26; i2++) {
        for (int i3 = i2 + 1; i3 < 26; i3++) {
          for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) {
              if (a[i][j - 1] - 'a' == i1 or a[i][j - 1] - 'a' == i2 or
                  a[i][j - 1] - 'a' == i3)
                mark[i][j] = 1;
              else
                mark[i][j] = 0;
              dis[i][j] = 1e9;
            }
          bfs();
        }
      }
    }
  }
  if (k == 2) {
    for (int i1 = 0; i1 < 26; i1++) {
      for (int i2 = i1 + 1; i2 < 26; i2++) {
        for (int i = 1; i <= n; i++)
          for (int j = 1; j <= m; j++) {
            if (a[i][j - 1] - 'a' == i1 or a[i][j - 1] - 'a' == i2)
              mark[i][j] = 1;
            else
              mark[i][j] = 0;
            dis[i][j] = 1e9;
          }
        bfs();
      }
    }
  }
  if (k == 1) {
    for (int i1 = 0; i1 < 26; i1++) {
      for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
          if (a[i][j - 1] - 'a' == i1)
            mark[i][j] = 1;
          else
            mark[i][j] = 0;
          dis[i][j] = 1e9;
        }
      bfs();
    }
  }
  if (d == 1e9) return cout << -1, 0;
  for (int i = 0; i < ans.size() - 1; i++) cout << ans[i];
}
