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, s, t, l = 2500, X, Y;
int u[30], a[2505], c, p[5], dis[2505];
string ans = "-1";
vector<int> v[2505];
int q[2505], qs, ps;
pair<pair<int, int>, int> P[2505];
string r[2505];
bool b[2505];
bool g[2505];
void cty() {
  q[1] = s;
  r[s] = "";
  qs = 1;
  memset(g, 0, sizeof(g));
  b[1] = 1;
  g[s] = 1;
  while (qs) {
    c = 0;
    ps = 0;
    for (int i = 1; i <= qs; i++) {
      if (b[i]) c++;
      for (int j = 0; j < v[q[i]].size(); j++) {
        int k = v[q[i]][j];
        if (g[k] || !u[a[k]]) continue;
        g[k] = 1;
        P[++ps] = make_pair(make_pair(c, a[k]), k);
        r[k] = r[q[i]] + (char)(96 + a[k]);
      }
    }
    sort(P + 1, P + ps + 1);
    b[1] = 1;
    for (int i = 2; i <= ps; i++) b[i] = (P[i].first != P[i - 1].first);
    qs = ps;
    for (int i = 1; i <= ps; i++) q[i] = P[i].second;
  }
  if (!g[t]) return;
  r[t].erase(r[t].size() - 1);
  if (r[t].size() < l || (r[t].size() == l && r[t] < ans))
    ans = r[t], l = ans.size();
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      c++;
      char x;
      scanf(" %c", &x);
      if (x == 'S')
        s = c;
      else if (x == 'T')
        t = c, X = i, Y = j;
      else
        a[c] = x - 'a' + 1;
      if (j < m) v[c].push_back(c + 1);
      if (i > 1) v[c].push_back(c - m);
      if (i < n) v[c].push_back(c + m);
      if (j > 1) v[c].push_back(c - 1);
    }
  }
  for (int i = 1; i <= k; i++) p[i] = i;
  while (!p[0]) {
    memset(u, 0, sizeof(u));
    u[0] = 1;
    for (int i = 1; i <= k; i++) u[p[i]] = 1;
    cty();
    int o = k;
    while (p[o] == 26 + o - k) o--;
    p[o]++;
    for (int i = o + 1; i <= k; i++) p[i] = p[i - 1] + 1;
  }
  cout << ans << endl;
  return 0;
}
