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 n, m, k;
int sx, sy, tx, ty;
char a[60][60];
bool ok[26];
int dis[60][60];
int d[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
int val[60][60];
pair<int, int> lst[60][60];
vector<pair<int, int> > tot[110];
vector<int> fst[26];
vector<int> snd[2510];
string ans;
bool good;
void solve() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) dis[i][j] = -1;
  dis[sx][sy] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(sx, sy));
  while (!q.empty()) {
    int x = q.front().first, y = q.front().second;
    q.pop();
    for (int i = 0; i < 4; i++) {
      int xx = x + d[i][0], yy = y + d[i][1];
      if (xx < 1 || xx > n || yy < 1 || yy > m) continue;
      if (dis[xx][yy] != -1 || (a[xx][yy] != 'T' && !ok[a[xx][yy] - 'a']))
        continue;
      dis[xx][yy] = dis[x][y] + 1;
      q.push(make_pair(xx, yy));
    }
  }
  if (dis[tx][ty] == -1) return;
  for (int i = 1; i <= dis[tx][ty]; i++) tot[i].clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (dis[i][j] <= 0 || dis[i][j] > dis[tx][ty]) continue;
      tot[dis[i][j]].push_back(make_pair(i, j));
    }
  }
  val[sx][sy] = 1;
  int sz = 1;
  for (int i = 1; i <= dis[tx][ty]; i++) {
    for (int j = 0; j < 26; j++) fst[j].clear();
    for (int j = 1; j <= sz; j++) snd[j].clear();
    for (int j = 0; j < (int)tot[i].size(); j++) {
      int x = tot[i][j].first, y = tot[i][j].second;
      if (a[x][y] != 'T') fst[a[x][y] - 'a'].push_back(j);
      val[x][y] = 1e9;
      for (int t = 0; t < 4; t++) {
        int xx = x + d[t][0], yy = y + d[t][1];
        if (xx < 1 || xx > n || yy < 1 || yy > m) continue;
        if (dis[xx][yy] != dis[x][y] - 1) continue;
        if (val[xx][yy] < val[x][y])
          val[x][y] = val[xx][yy], lst[x][y] = make_pair(xx, yy);
      }
    }
    for (int j = 0; j < 26; j++) {
      for (int t = 0; t < (int)fst[j].size(); t++) {
        int x = tot[i][fst[j][t]].first, y = tot[i][fst[j][t]].second;
        snd[val[x][y]].push_back(fst[j][t]);
      }
    }
    int tsz = 0;
    int lx = 0, ly = 0, lv = 0;
    for (int j = 1; j <= sz; j++) {
      for (int t = 0; t < (int)snd[j].size(); t++) {
        int x = tot[i][snd[j][t]].first, y = tot[i][snd[j][t]].second;
        if (lx && ly && lv == val[x][y] && a[lx][ly] == a[x][y])
          lv = val[x][y], val[x][y] = tsz;
        else
          lv = val[x][y], val[x][y] = ++tsz;
        lx = x;
        ly = y;
      }
    }
    sz = tsz;
  }
  string res = "";
  int x = lst[tx][ty].first, y = lst[tx][ty].second;
  while (x != sx || y != sy) {
    res = a[x][y] + res;
    int ttx = lst[x][y].first, tty = lst[x][y].second;
    x = ttx;
    y = tty;
  }
  if (!good || (int)res.size() < (int)ans.size() ||
      ((int)res.size() == (int)ans.size() && res < ans))
    ans = res;
  good = true;
}
void dfs(int x, int lst) {
  if (x > k) {
    solve();
    return;
  }
  for (int i = lst; i < 26; i++) {
    ok[i] = true;
    dfs(x + 1, i + 1);
    ok[i] = false;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%s", a[i] + 1);
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == 'S') sx = i, sy = j;
      if (a[i][j] == 'T') tx = i, ty = j;
    }
  }
  dfs(1, 0);
  if (!good)
    printf("-1");
  else
    cout << ans;
  return 0;
}
