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 X[] = {-1, 0, 1, 0}, Y[] = {0, 1, 0, -1}, K = 'z' - 'a' + 2;
char a[52][52];
queue<pair<int, int> > q;
int n, m, k, mini = (1 << 30), sx, sy, fx, fy;
string mn;
vector<char> p;
set<char> buk;
vector<char> b;
bool chk(int i, int j) {
  return (i >= 0 && i < n && j >= 0 && j < m &&
          (a[i][j] == 'S' || a[i][j] == 'T' || (int)mn.find(a[i][j]) != -1));
}
void bfs(int x1, int y1, int x2, int y2) {
  vector<char> path;
  path.clear();
  int d[52][52], pr[52][52];
  bool u[52][52] = {false};
  vector<vector<pair<int, int> > > qe[K];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) d[i][j] = (1 << 30);
  d[x2][y2] = 0;
  q.push(make_pair(x2, y2));
  int nx, ny, x, y;
  pair<int, int> v;
  while (!q.empty()) {
    v = q.front();
    x = v.first;
    y = v.second;
    q.pop();
    for (int c = 0; c < 4; c++) {
      nx = x + X[c];
      ny = y + Y[c];
      if (chk(nx, ny) && d[nx][ny] == (1 << 30)) {
        d[nx][ny] = d[x][y] + 1;
        q.push(make_pair(nx, ny));
      }
    }
  }
  if (d[x1][y1] == (1 << 30)) return;
  pr[x1][y1] = -1;
  u[x1][y1] = true;
  for (int i = 0; i < K; i++) qe[i].resize(d[x1][y1]);
  qe[K - 1][d[x1][y1] - 1].push_back(make_pair(x1, y1));
  int minc = K - 1, fminc, newminc;
  for (int i = d[x1][y1] - 1; i >= 0; i--) {
    newminc = K;
    for (int j = 0; j < qe[minc][i].size(); j++) {
      fminc = K - 1;
      x = qe[minc][i][j].first;
      y = qe[minc][i][j].second;
      for (int c = 0; c < 4; c++) {
        nx = x + X[c];
        ny = y + Y[c];
        if (chk(nx, ny) && a[nx][ny] != 'T' && d[nx][ny] < d[x][y])
          fminc = min(fminc, (int)a[nx][ny] - 'a');
      }
      for (int c = 0; c < 4; c++) {
        nx = x + X[c];
        ny = y + Y[c];
        if (a[nx][ny] == 'T') {
          pr[x2][y2] = c;
          nx = x2 - X[pr[x2][y2]];
          ny = y2 - Y[pr[x2][y2]];
          while (pr[nx][ny] != -1) {
            path.push_back(a[nx][ny]);
            x2 = nx, y2 = ny;
            nx = x2 - X[pr[x2][y2]];
            ny = y2 - Y[pr[x2][y2]];
          }
          reverse(path.begin(), path.end());
          if (d[x1][y1] < mini || d[x1][y1] == mini && path < p) {
            p = path;
            mini = d[x1][y1];
          }
          return;
        }
        if (chk(nx, ny) && a[nx][ny] - 'a' == fminc && d[nx][ny] < d[x][y] &&
            !u[nx][ny]) {
          u[nx][ny] = true;
          pr[nx][ny] = c;
          qe[fminc][i - 1].push_back(make_pair(nx, ny));
        }
      }
      newminc = min(newminc, fminc);
    }
    minc = newminc;
  }
}
void gen(int r, int ind) {
  if (r == 0) {
    bfs(sx, sy, fx, fy);
    return;
  } else if (ind >= (int)b.size())
    return;
  else {
    mn = mn + b[ind];
    gen(r - 1, ind + 1);
    mn.erase(mn.size() - 1, 1);
    gen(r, ind + 1);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      if (a[i][j] == 'S')
        sx = i, sy = j;
      else if (a[i][j] == 'T')
        fx = i, fy = j;
      else
        buk.insert(a[i][j]);
    }
  while (buk.size()) {
    b.push_back((*buk.begin()));
    buk.erase(buk.begin());
  }
  gen(min(k, (int)b.size()), 0);
  if (mini == (1 << 30))
    cout << -1;
  else
    for (int i = 0; i < (int)p.size(); i++) cout << p[i];
  return 0;
}
