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>
#pragma comment(linker, "/STACK:10000000")
using namespace std;
template <typename T>
inline T abs(T a) {
  return ((a < 0) ? -a : a);
}
template <typename T>
inline T sqr(T a) {
  return a * a;
}
const int INF = (int)1E9 + 7;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int NMAX = 52;
char a[NMAX][NMAX];
int n, m, k;
pair<int, int> s, t;
bool can[300];
const string FAIL = "$";
string ans = FAIL;
void upd(const string& s) {
  if (s == FAIL) return;
  if (ans == FAIL)
    ans = s;
  else {
    if (int((ans).size()) > int((s).size()) ||
        (int((ans).size()) == int((s).size()) && ans > s))
      ans = s;
  }
}
int d1[NMAX][NMAX], d2[NMAX][NMAX];
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
inline bool in(const pair<int, int>& nv) {
  return !(nv.first < 0 || nv.first >= n || nv.second < 0 || nv.second >= m);
}
void genD(pair<int, int> s, int d[NMAX][NMAX]) {
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) d[i][j] = INF;
  queue<pair<int, int> > q;
  q.push(s), d[s.first][s.second] = 0;
  while (!q.empty()) {
    pair<int, int> v = q.front();
    q.pop();
    for (int i = 0; i < int(4); ++i) {
      pair<int, int> nv(v.first + dx[i], v.second + dy[i]);
      if (!in(nv)) continue;
      if (!can[a[nv.first][nv.second]]) continue;
      if (d[nv.first][nv.second] > d[v.first][v.second] + 1) {
        d[nv.first][nv.second] = d[v.first][v.second] + 1;
        q.push(nv);
      }
    }
  }
}
string get() {
  genD(s, d1);
  genD(t, d2);
  if (d1[t.first][t.second] > INF / 2) return FAIL;
  string ans;
  vector<pair<int, int> > q;
  q.push_back(s);
  for (int step = 0; step < int(d1[t.first][t.second]); ++step) {
    vector<pair<int, int> > nq;
    char minC = -1;
    for (int i = 0; i < int(int((q).size())); ++i) {
      pair<int, int> v = q[i];
      for (int j = 0; j < int(4); ++j) {
        pair<int, int> nv(v.first + dx[j], v.second + dy[j]);
        if (!in(nv) || !can[a[nv.first][nv.second]]) continue;
        if (d1[nv.first][nv.second] + d2[nv.first][nv.second] !=
            d1[t.first][t.second])
          continue;
        if (d1[nv.first][nv.second] <= d1[v.first][v.second]) continue;
        if (minC == -1) minC = a[nv.first][nv.second];
        minC = min(minC, a[nv.first][nv.second]);
      }
    }
    ans += minC;
    for (int i = 0; i < int(int((q).size())); ++i) {
      pair<int, int> v = q[i];
      for (int j = 0; j < int(4); ++j) {
        pair<int, int> nv(v.first + dx[j], v.second + dy[j]);
        if (!in(nv) || !can[a[nv.first][nv.second]]) continue;
        if (d1[nv.first][nv.second] + d2[nv.first][nv.second] !=
            d1[t.first][t.second])
          continue;
        if (d1[nv.first][nv.second] <= d1[v.first][v.second]) continue;
        if (a[nv.first][nv.second] != minC) continue;
        nq.push_back(nv);
      }
    }
    sort((nq).begin(), (nq).end());
    nq.erase(unique((nq).begin(), (nq).end()), nq.end());
    q = nq;
  }
  assert(int((q).size()) == 1 && q[0] == t);
  ans.erase(int((ans).size()) - 1);
  return ans;
}
void brute(char c, int cnt) {
  if (cnt > k) return;
  if (cnt == k) {
    upd(get());
    return;
  }
  if (c > 'z') return;
  can[c] = true;
  brute(c + 1, cnt + 1);
  can[c] = false;
  brute(c + 1, cnt);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) {
      scanf(" %c ", &a[i][j]);
      if (a[i][j] == 'S') s = pair<int, int>(i, j);
      if (a[i][j] == 'T') t = pair<int, int>(i, j);
    }
  can['S'] = true, can['T'] = true;
  brute('a', 0);
  if (ans == FAIL)
    puts("-1");
  else
    cout << ans << endl;
  return 0;
}
