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;
template <class T>
T abs(T x) {
  return x > 0 ? x : (-x);
}
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 57;
char ty[maxn][maxn];
bool allow[300];
int d[maxn][maxn];
int f[maxn][maxn];
int n, m, k;
int sx, sy, fx, fy;
int Q[maxn * maxn * 2];
int qs, qt;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
void bfs() {
  d[sx][sy] = 0;
  qs = 0, qt = 0;
  Q[qs++] = sx * maxn + sy;
  while (qs > qt) {
    int x = Q[qt] / maxn;
    int y = Q[qt] % maxn;
    if (x == fx && y == fy) break;
    ++qt;
    for (int dir = 0; dir < 4; ++dir) {
      int X = x + dx[dir], Y = y + dy[dir];
      if (X < 0 || Y < 0 || X >= n || Y >= m) continue;
      if (d[X][Y] != -1) continue;
      if (!allow[ty[X][Y]]) continue;
      d[X][Y] = d[x][y] + 1;
      f[X][Y] = x * maxn + y;
      Q[qs++] = X * maxn + Y;
    }
  }
}
string cur;
string rf[maxn][maxn];
bool uf[maxn][maxn];
void rest() {
  memset(uf, 0, sizeof(uf));
  rf[sx][sy] = "S";
  qs = 0, qt = 0;
  uf[sx][sy] = true;
  Q[qs++] = sx * maxn + sy;
  while (qs > qt) {
    int x = Q[qt] / maxn;
    int y = Q[qt] % maxn;
    if (x == fx && y == fy) break;
    ++qt;
    string& s = rf[x][y];
    for (int dir = 0; dir < 4; ++dir) {
      int X = x + dx[dir], Y = y + dy[dir];
      if (X < 0 || Y < 0 || X >= n || Y >= m) continue;
      if (d[X][Y] != d[x][y] + 1) continue;
      if (!uf[X][Y]) {
        uf[X][Y] = true;
        rf[X][Y] = rf[x][y] + ty[X][Y];
        Q[qs++] = X * maxn + Y;
      } else {
        rf[X][Y] = min(rf[X][Y], rf[x][y] + ty[X][Y]);
      }
    }
  }
  cur = rf[fx][fy];
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < m; ++j) {
      ty[i][j] = s[j];
      if (islower(s[j])) {
        continue;
      }
      if (s[j] == 'S')
        sx = i, sy = j;
      else
        fx = i, fy = j;
    }
  }
  int res = 100500100;
  string rs;
  bool rsf = false;
  set<long long> S;
  for (int i = 'a'; i <= 'z'; ++i)
    for (int j = 'a'; j <= i; ++j)
      for (int p = 'a'; p <= j; ++p)
        for (int q = 'a'; q <= p; ++q) {
          long long h = 123;
          if (k >= 1) h = h * 2184057 + i;
          if (k >= 2) h = h * 2184057 + j;
          if (k >= 3) h = h * 2184057 + p;
          if (k >= 4) h = h * 2184057 + q;
          if (S.count(h)) continue;
          S.insert(h);
          memset(allow, 0, sizeof(allow));
          if (k >= 1) allow[i] = true;
          if (k >= 2) allow[j] = true;
          if (k >= 3) allow[p] = true;
          if (k >= 4) allow[q] = true;
          allow['S'] = allow['T'] = true;
          memset(d, -1, sizeof(d));
          bfs();
          if (d[fx][fy] == -1) continue;
          if (d[fx][fy] > res) continue;
          if (res > d[fx][fy]) res = d[fx][fy];
        }
  S.clear();
  int bad = 1000000;
  for (int i = 'a'; i <= 'z'; ++i)
    for (int j = 'a'; j <= i; ++j)
      for (int p = 'a'; p <= j; ++p)
        for (int q = 'a'; q <= p; ++q) {
          if (i >= bad && j >= bad && p >= bad && q >= bad) continue;
          long long h = 123;
          if (k >= 1) h = h * 2184057 + i;
          if (k >= 2) h = h * 2184057 + j;
          if (k >= 3) h = h * 2184057 + p;
          if (k >= 4) h = h * 2184057 + q;
          if (S.count(h)) continue;
          S.insert(h);
          memset(allow, 0, sizeof(allow));
          if (k >= 1) allow[i] = true;
          if (k >= 2) allow[j] = true;
          if (k >= 3) allow[p] = true;
          if (k >= 4) allow[q] = true;
          allow['S'] = allow['T'] = true;
          memset(d, -1, sizeof(d));
          bfs();
          if (d[fx][fy] == -1) continue;
          if (d[fx][fy] > res) continue;
          rest();
          if (!rsf || cur < rs) {
            rsf = true, rs = cur;
            bad = -1;
            for (int t = 0; t < ((int)(cur).size()); ++t)
              bad = max(bad, (int)cur[t]);
          }
        }
  if (res == 100500100) res = -1;
  if (res == -1)
    cout << res << "\n";
  else
    cout << rs.substr(1, ((int)(rs).size()) - 2) << "\n";
  return 0;
}
