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;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
pair<int, int> st, en;
int m, n, k;
char a[55][55];
bool b[55][55];
int d[55][55];
string Z[55][55];
bool inq[55][55];
int solve1() {
  queue<pair<int, int> > q;
  q.push(st);
  memset(d, 0x1f, sizeof(d));
  d[st.first][st.second] = 0;
  while (!q.empty()) {
    int u = q.front().first;
    int v = q.front().second;
    q.pop();
    for (int h = 0; h < (4); ++h) {
      int x = u + dx[h];
      int y = v + dy[h];
      if (0 <= x && x < m && 0 <= y && y < n && b[x][y]) {
        if (d[x][y] > d[u][v] + 1) {
          d[x][y] = d[u][v] + 1;
          q.push(make_pair(x, y));
        }
      }
    }
  }
  return d[en.first][en.second];
}
string solve2() {
  queue<pair<int, int> > q;
  memset(inq, 0, sizeof(inq));
  q.push(st);
  inq[st.first][st.second] = true;
  Z[st.first][st.second] = "";
  while (!q.empty()) {
    int u = q.front().first;
    int v = q.front().second;
    q.pop();
    for (int h = 0; h < (4); ++h) {
      int x = u + dx[h];
      int y = v + dy[h];
      if (0 <= x && x < m && 0 <= y && y < n && b[x][y]) {
        if (d[x][y] == d[u][v] + 1) {
          if (!inq[x][y]) {
            q.push(make_pair(x, y));
            inq[x][y] = true;
            Z[x][y] = Z[u][v] + a[x][y];
          } else if (Z[x][y] > Z[u][v] + a[x][y])
            Z[x][y] = Z[u][v] + a[x][y];
        }
      }
    }
  }
  return Z[en.first][en.second];
}
int main() {
  scanf("%d%d%d", &m, &n, &k);
  gets(a[0]);
  for (int i = 0; i < (m); ++i) gets(a[i]);
  for (int i = 0; i < (m); ++i)
    for (int j = 0; j < (n); ++j)
      if (a[i][j] == 'S')
        st = make_pair(i, j);
      else if (a[i][j] == 'T')
        en = make_pair(i, j);
  int res = 50 * 50 * 50;
  for (int c1 = 0; c1 < (26); ++c1)
    for (int c2 = 0; c2 < (c1 + 1); ++c2)
      for (int c3 = 0; c3 < (c2 + 1); ++c3)
        for (int c4 = 0; c4 < (c3 + 1); ++c4) {
          set<int> second;
          second.insert(c1);
          second.insert(c2);
          second.insert(c3);
          second.insert(c4);
          if (second.size() == k) {
            memset(b, 0, sizeof(b));
            for (int i = 0; i < (m); ++i)
              for (int j = 0; j < (n); ++j)
                if (a[i][j] == 'S' || a[i][j] == 'T' ||
                    second.count(a[i][j] - 'a'))
                  b[i][j] = true;
            int len = solve1();
            res = min(res, len);
          }
        }
  if (res == 50 * 50 * 50)
    cout << -1 << endl;
  else {
    string beststr = "";
    for (int c1 = 0; c1 < (26); ++c1)
      for (int c2 = 0; c2 < (c1 + 1); ++c2)
        for (int c3 = 0; c3 < (c2 + 1); ++c3)
          for (int c4 = 0; c4 < (c3 + 1); ++c4) {
            set<int> second;
            second.insert(c1);
            second.insert(c2);
            second.insert(c3);
            second.insert(c4);
            if (second.size() == k) {
              memset(b, 0, sizeof(b));
              for (int i = 0; i < (m); ++i)
                for (int j = 0; j < (n); ++j)
                  if (a[i][j] == 'S' || a[i][j] == 'T' ||
                      second.count(a[i][j] - 'a'))
                    b[i][j] = true;
              int len = solve1();
              if (len == res) {
                string path = solve2();
                if (beststr == "" || beststr > path) beststr = path;
              }
            }
          }
    beststr = beststr.substr(0, beststr.length() - 1);
    cout << beststr << endl;
  }
  return 0;
}
