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;
int xs, ys, xt, yt;
int r, c, k, dist[55][55], ans, ok[30];
char mat[55][55];
bool help[55][55];
string str;
void solve() {
  queue<pair<int, int> > q;
  memset(dist, 0x3f3f3f3f, sizeof(dist));
  dist[xt][yt] = 0;
  q.push(pair<int, int>(xt, yt));
  while (!q.empty()) {
    int i = q.front().first, j = q.front().second;
    q.pop();
    if (i == xs && j == ys) break;
    if (i && dist[i - 1][j] == 0x3f3f3f3f &&
        (mat[i - 1][j] == 'S' || ok[mat[i - 1][j] - 'a'])) {
      dist[i - 1][j] = dist[i][j] + 1;
      q.push(pair<int, int>(i - 1, j));
    }
    if (i < r - 1 && dist[i + 1][j] == 0x3f3f3f3f &&
        (mat[i + 1][j] == 'S' || ok[mat[i + 1][j] - 'a'])) {
      dist[i + 1][j] = dist[i][j] + 1;
      q.push(pair<int, int>(i + 1, j));
    }
    if (j && dist[i][j - 1] == 0x3f3f3f3f &&
        (mat[i][j - 1] == 'S' || ok[mat[i][j - 1] - 'a'])) {
      dist[i][j - 1] = dist[i][j] + 1;
      q.push(pair<int, int>(i, j - 1));
    }
    if (j < c - 1 && dist[i][j + 1] == 0x3f3f3f3f &&
        (mat[i][j + 1] == 'S' || ok[mat[i][j + 1] - 'a'])) {
      dist[i][j + 1] = dist[i][j] + 1;
      q.push(pair<int, int>(i, j + 1));
    }
  }
  if (dist[xs][ys] == 0x3f3f3f3f || dist[xs][ys] > ans) return;
  vector<pair<int, int> > curr, next;
  char best, ch;
  string s;
  curr.push_back(pair<int, int>(xs, ys));
  memset(help, false, sizeof(help));
  for (int x = dist[xs][ys]; x > 0; x--) {
    best = 'z' + 1;
    for (int A = 0; A < curr.size(); A++) {
      int i = curr[A].first, j = curr[A].second;
      if (i && dist[i - 1][j] == x - 1 && !help[i - 1][j]) {
        ch = mat[i - 1][j];
        if (ch < best) {
          best = ch;
          next.clear();
          next.push_back(pair<int, int>(i - 1, j));
          help[i - 1][j] = true;
        } else if (ch == best && !help[i - 1][j]) {
          next.push_back(pair<int, int>(i - 1, j));
          help[i - 1][j] = true;
        }
      }
      if (i < r - 1 && dist[i + 1][j] == x - 1 && !help[i + 1][j]) {
        ch = mat[i + 1][j];
        if (ch < best) {
          best = ch;
          next.clear();
          next.push_back(pair<int, int>(i + 1, j));
          help[i + 1][j] = true;
        } else if (ch == best && !help[i + 1][j]) {
          next.push_back(pair<int, int>(i + 1, j));
          help[i + 1][j] = true;
        }
      }
      if (j && dist[i][j - 1] == x - 1 && !help[i][j - 1]) {
        ch = mat[i][j - 1];
        if (ch < best) {
          best = ch;
          next.clear();
          next.push_back(pair<int, int>(i, j - 1));
          help[i][j - 1] = true;
        } else if (ch == best && !help[i][j - 1]) {
          next.push_back(pair<int, int>(i, j - 1));
          help[i][j - 1] = true;
        }
      }
      if (j < c - 1 && dist[i][j + 1] == x - 1 && !help[i][j + 1]) {
        ch = mat[i][j + 1];
        if (ch < best) {
          best = ch;
          next.clear();
          next.push_back(pair<int, int>(i, j + 1));
          help[i][j + 1] = true;
        } else if (ch == best && !help[i][j + 1]) {
          next.push_back(pair<int, int>(i, j + 1));
          help[i][j + 1] = true;
        }
      }
    }
    if (best != 'T') s.push_back(best);
    curr.clear();
    curr = next;
  }
  if (dist[xs][ys] < ans) {
    ans = dist[xs][ys];
    str = s;
  } else if (dist[xs][ys] == ans && s < str)
    str = s;
  return;
}
void go(int x, int last) {
  if (x == k) {
    solve();
    return;
  }
  for (int i = last + 1; i < 26; i++)
    if (!ok[i]) {
      ok[i] = 1;
      go(x + 1, i);
      ok[i] = 0;
    }
  return;
}
int main() {
  cin >> r >> c >> k;
  for (int i = 0; i < r; i++) cin >> mat[i];
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) {
      if (mat[i][j] == 'S') {
        xs = i;
        ys = j;
      } else if (mat[i][j] == 'T') {
        xt = i;
        yt = j;
      }
    }
  ans = 0x3f3f3f3f;
  memset(ok, 0, sizeof(ok));
  go(0, -1);
  if (ans == 0x3f3f3f3f)
    cout << -1;
  else
    cout << str;
  cout << endl;
  return 0;
}