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;
char a[55][55];
vector<vector<int> > c;
int was[26];
int wass[55][55];
pair<int, int> Par[55][55];
int x[2510];
int y[2510];
int par[2510];
int d[2510];
int D[55][55];
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
string Path[55][55];
void dfs(int p, int k, vector<int> a) {
  if (k == 0) {
    c.push_back(a);
    return;
  }
  if (p == 26) {
    if (a.size()) c.push_back(a);
    return;
  }
  dfs(p + 1, k, a);
  a.push_back(p);
  dfs(p + 1, k - 1, a);
}
int main() {
  vector<int> aa;
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  dfs(0, k, aa);
  for (int i = (0); i < (n); ++i) scanf("%s", &a[i]);
  int x1, y1, x2, y2;
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == 'S') {
        x1 = i;
        y1 = j;
      } else if (a[i][j] == 'T') {
        x2 = i;
        y2 = j;
      }
  bool f = false;
  string res = "";
  for (int it = (0); it < (c.size()); ++it) {
    memset(wass, 0, sizeof(wass));
    for (int i = (0); i < (n); ++i)
      for (int j = (0); j < (m); ++j) {
        D[i][j] = 2000000000;
      }
    if (it)
      for (int i = (0); i < (c[it - 1].size()); ++i) was[c[it - 1][i]] = 0;
    for (int i = (0); i < (c[it].size()); ++i) was[c[it][i]] = 1;
    int l = 0, r = 0;
    x[0] = x2;
    y[0] = y2;
    d[0] = 0;
    wass[x2][y2] = 1;
    par[0] = -1;
    int v = -1;
    D[x2][y2] = 0;
    while (l <= r) {
      vector<pair<char, pair<int, int> > > can;
      if (a[x[l]][y[l]] == 'S') v = 1;
      for (int k = (0); k < (4); ++k) {
        int xx = x[l] + dx[k];
        int yy = y[l] + dy[k];
        if ((xx > -1) && (yy > -1) && (xx < n) && (yy < m) &&
            ((a[xx][yy] == 'S') || (was[a[xx][yy] - 'a'] == 1))) {
          if (wass[xx][yy] == 0) {
            can.push_back(make_pair(a[xx][yy], make_pair(xx, yy)));
          }
        }
      }
      sort(can.begin(), can.end());
      for (int i = (0); i < (can.size()); ++i) {
        int xx = can[i].second.first;
        int yy = can[i].second.second;
        wass[xx][yy] = 1;
        r++;
        x[r] = xx;
        y[r] = yy;
        d[r] = d[l] + 1;
        D[xx][yy] = d[r];
        par[r] = l;
      }
      l++;
    }
    if (v != -1) {
      vector<pair<int, int> > cur;
      cur.push_back(make_pair(x1, y1));
      string res1 = "";
      memset(wass, 0, sizeof(wass));
      while (1) {
        int min1 = 2000000000;
        bool ff = true;
        for (int i = (0); i < (cur.size()); ++i)
          if ((cur[i].first == x2) && (cur[i].second == y2)) ff = false;
        if (!ff) break;
        for (int i = (0); i < (cur.size()); ++i)
          for (int j = (0); j < (4); ++j) {
            int xx = cur[i].first + dx[j];
            int yy = cur[i].second + dy[j];
            if ((xx > -1) && (yy > -1) && (xx < n) && (yy < m) &&
                (wass[xx][yy] == 0) &&
                (D[xx][yy] == D[cur[i].first][cur[i].second] - 1)) {
              min1 = ((min1 < a[xx][yy] - 'a') ? (min1) : (a[xx][yy] - 'a'));
            }
          }
        vector<pair<int, int> > cur1;
        for (int i = (0); i < (cur.size()); ++i)
          for (int j = (0); j < (4); ++j) {
            int xx = cur[i].first + dx[j];
            int yy = cur[i].second + dy[j];
            if ((xx > -1) && (yy > -1) && (xx < n) && (yy < m) &&
                (wass[xx][yy] == 0) &&
                (D[xx][yy] == D[cur[i].first][cur[i].second] - 1) &&
                (a[xx][yy] == 'a' + min1)) {
              wass[xx][yy] = 1;
              cur1.push_back(make_pair(xx, yy));
            }
          }
        cur = cur1;
        res1 += 'a' + min1;
      }
      res1 = res1.substr(0, res1.size() - 1);
      if (!f) {
        f = true;
        res = res1;
      } else if (res1.size() < res.size())
        res = res1;
      else if ((res1.size() == res.size()) && (res1 < res))
        res = res1;
    }
  }
  if (!f)
    printf("-1\n");
  else
    cout << res << endl;
  return 0;
}
