Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 64;
const int inf = 0x3f3f3f3f;
int n, m, k;
int sx, sy, tx, ty;
char mat[maxn][maxn];
bool allow[128];
struct point {
  int x, y, pre;
  point(int xx = 0, int yy = 0, int pp = 0) : x(xx), y(yy), pre(pp) {}
  friend bool operator<(const point& lhs, const point& rhs) {
    return (lhs.pre != rhs.pre) ? lhs.pre < rhs.pre
                                : mat[lhs.x][lhs.y] < mat[rhs.x][rhs.y];
  }
};
int prex[maxn][maxn];
int prey[maxn][maxn];
int dist[maxn][maxn];
int dx[4] = {-1, 0, 0, 1};
int dy[4] = {0, -1, 1, 0};
void bfs() {
  memset(dist, 63, sizeof(dist));
  vector<point> U, V;
  dist[sx][sy] = 0;
  U.push_back(point(sx, sy, 0));
  int cnt = 0;
  while (!U.empty()) {
    sort(U.begin(), U.end());
    for (int i = 0, mxRank = 0; i < U.size(); ++i) {
      point u = U[i];
      if (i != 0 && (mat[U[i].x][U[i].y] != mat[U[i - 1].x][U[i - 1].y] ||
                     U[i].pre != U[i - 1].pre))
        ++mxRank;
      for (int j = 0; j < 4; ++j) {
        point v(u.x + dx[j], u.y + dy[j], mxRank);
        if (dist[v.x][v.y] == inf && allow[mat[v.x][v.y]]) {
          V.push_back(v);
          dist[v.x][v.y] = dist[u.x][u.y] + 1;
          prex[v.x][v.y] = u.x;
          prey[v.x][v.y] = u.y;
        }
        if (v.x == tx && v.y == ty) return;
      }
    }
    U.swap(V);
    V.clear();
  }
}
int main() {
  int ans = inf;
  vector<string> s;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cin >> mat[i][j];
      if (mat[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (mat[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  allow['T'] = true;
  for (char l1 = 'a'; l1 <= 'z'; ++l1) {
    for (char l2 = l1; l2 <= 'z'; ++l2) {
      for (char l3 = l2; l3 <= 'z'; ++l3) {
        for (char l4 = l3; l4 <= 'z'; ++l4) {
          int cnt = 1;
          if (l2 != l1) ++cnt;
          if (l3 != l2) ++cnt;
          if (l4 != l3) ++cnt;
          if (cnt > k) continue;
          allow[l1] = allow[l2] = allow[l3] = allow[l4] = true;
          bfs();
          allow[l1] = allow[l2] = allow[l3] = allow[l4] = false;
          if (dist[tx][ty] > ans || dist[tx][ty] == inf) continue;
          if (dist[tx][ty] < ans) {
            ans = dist[tx][ty];
            s.clear();
          }
          int fx = tx, fy = ty, ffx, ffy;
          string tmps;
          while (true) {
            ffx = prex[fx][fy];
            ffy = prey[fx][fy];
            fx = ffx;
            fy = ffy;
            if (fx == sx && fy == sy) break;
            tmps = mat[fx][fy] + tmps;
          }
          s.push_back(tmps);
        }
      }
    }
  }
  if (ans != inf) {
    sort(s.begin(), s.end());
    for (int i = 0; i < s[0].size(); ++i) cout << (s[0][i]);
    cout << endl;
  } else {
    cout << EOF << endl;
  }
}