Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int n, m, k;
string a[55];
string ans;
int dist = inf;
bool can[55][55];
vector<char> v;
bool ok(int x, int y) { return x >= 0 && x < n && y >= 0 && y < m; }
void fill(vector<char> &v) {
  map<char, bool> was;
  for (int i = 0; i < v.size(); i++) was[v[i]] = 1;
  was['S'] = 1;
  was['T'] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      can[i][j] = was[a[i][j]];
    }
  }
}
int level[55][55];
int d[55][55];
string solve() {
  memset(d, -1, sizeof d);
  queue<pair<int, int> > q;
  int endx = -1, endy = -1;
  int startx = -1, starty = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'T') {
        q.push(make_pair(i, j));
        d[i][j] = 0;
        startx = i;
        starty = j;
      }
      if (a[i][j] == 'S') {
        endx = i;
        endy = j;
      }
    }
  }
  while (!q.empty()) {
    pair<int, int> p = q.front();
    q.pop();
    int x = p.first;
    int y = p.second;
    for (int i = 0; i < 4; i++) {
      int nx = x + dx[i];
      int ny = y + dy[i];
      if (ok(nx, ny) && can[nx][ny] && d[nx][ny] == -1) {
        q.push(make_pair(nx, ny));
        d[nx][ny] = d[x][y] + 1;
      }
    }
  }
  if (d[endx][endy] == -1) {
    string ans = "";
    ans += (char)('z' + 1);
    return ans;
  }
  string ans = "";
  int len = d[endx][endy];
  vector<pair<int, int> > cur;
  cur.push_back(make_pair(endx, endy));
  while (len) {
    vector<pair<char, pair<int, int> > > temp;
    for (int i = 0; i < cur.size(); i++) {
      int x = cur[i].first;
      int y = cur[i].second;
      for (int j = 0; j < 4; j++) {
        int nx = x + dx[j];
        int ny = y + dy[j];
        if (ok(nx, ny) && d[nx][ny] == len - 1) {
          temp.push_back(make_pair(a[nx][ny], make_pair(nx, ny)));
        }
      }
    }
    sort(temp.begin(), temp.end());
    temp.resize(unique(temp.begin(), temp.end()) - temp.begin());
    int en = 0;
    while (en < temp.size() && temp[en].first == temp[0].first) en++;
    vector<pair<int, int> > tmp;
    for (int i = 0; i < en; i++) tmp.push_back(temp[i].second);
    ans += temp[0].first;
    cur = tmp;
    len--;
  }
  ans.erase((int)ans.size() - 1);
  return ans;
}
void go(int ind, char last = (char)('a' - 1)) {
  if (ind == k) {
    fill(v);
    string cur = solve();
    if (cur[0] > 'z') return;
    if (cur.size() < dist) {
      dist = cur.size();
      ans = cur;
    } else if (cur.size() == dist) {
      ans = min(ans, cur);
    }
    return;
  }
  for (char x = last + 1; x <= 'z'; x++) {
    v.push_back(x);
    go(ind + 1, x);
    v.pop_back();
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  ans += (char)('z' + 1);
  go(0);
  if (ans[0] == (char)('z' + 1))
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}