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;
vector<string> mat;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
long long d(long long x1, long long x2, long long y1, long long y2) {
  return (abs(y2 - y1) + abs(x2 - x1));
}
int main() {
  long long n, m, k, sx, sy, gx, gy;
  cin >> n >> m >> k;
  string tmp;
  for (int i = 0; i < n; i++) {
    cin >> tmp;
    mat.push_back(tmp);
    for (int j = 0; j < m; j++) {
      if (mat[i][j] == 'S') {
        sy = i;
        sx = j;
      } else if (mat[i][j] == 'T') {
        gy = i;
        gx = j;
      }
    }
  }
  priority_queue<
      pair<long long, pair<pair<string, long long>, long long> >,
      vector<pair<long long, pair<pair<string, long long>, long long> > >,
      greater<pair<long long, pair<pair<string, long long>, long long> > > >
      pq;
  set<pair<long long, long long> > ss;
  pq.push(
      make_pair(d(sx, gx, sy, gy), make_pair(make_pair("", 0), sy * m + sx)));
  while (!pq.empty()) {
    long long an = pq.top().second.second;
    long long ay = pq.top().second.second / m;
    long long ax = pq.top().second.second % m;
    string astr = pq.top().second.first.first;
    long long mask = pq.top().second.first.second;
    pq.pop();
    if (ss.count(make_pair(mask, an))) continue;
    ss.insert(make_pair(mask, an));
    for (int i = 0; i < 4; i++) {
      int nx = ax + dx[i];
      int ny = ay + dy[i];
      if (ny < 0 || ny >= n || nx < 0 || nx >= m) continue;
      if (ny == gy && nx == gx) {
        cout << astr << endl;
        return 0;
      }
      string nstr = astr + mat[ny][nx];
      long long nmask = mask | (1 << (mat[ny][nx] - 'a'));
      if (__builtin_popcount(nmask) <= k)
        pq.push(make_pair(d(nx, gx, ny, gy) + nstr.size(),
                          make_pair(make_pair(nstr, nmask), ny * m + nx)));
    }
  }
  cout << "-1" << endl;
}
