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;
const int maxN = 55;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int a[27], n, m;
char s[maxN][maxN];
int d[maxN][maxN];
queue<pair<int, int> > q;
void f(int x, int y) {
  memset(d, 0xff, sizeof(d));
  d[x][y] = 0;
  assert(q.empty());
  q.push(make_pair(x, y));
  while (!q.empty()) {
    pair<int, int> t = q.front();
    q.pop();
    for (int i = 0; i < (4); ++i) {
      x = t.first + dx[i], y = t.second + dy[i];
      if ((0 <= (x) && (x) < (n)) && (0 <= (y) && (y) < (m)) &&
          (s[x][y] == 'S' || (s[x][y] != 'T' && a[s[x][y] - 'a'])) &&
          d[x][y] < 0) {
        d[x][y] = d[t.first][t.second] + 1;
        q.push(make_pair(x, y));
      }
    }
  }
}
vector<pair<int, int> > cur, nx;
string RES;
void reconstruct(int x, int y) {
  char ch;
  vector<pair<int, int> >().swap(nx);
  vector<pair<int, int> >().swap(cur);
  cur.push_back(make_pair(x, y));
  RES = "";
  while (1) {
    int dist = d[cur[0].first][cur[0].second];
    for (__typeof((cur).begin()) i = ((cur).begin()); i != (cur).end(); ++i)
      assert(dist == d[i->first][i->second]);
    if (dist == 1) return;
    assert(nx.empty() && (int)(cur).size() < maxN * maxN);
    ch = 'z' + 1;
    for (__typeof((cur).begin()) i = ((cur).begin()); i != (cur).end(); ++i)
      for (int j = 0; j < (4); ++j) {
        x = i->first + dx[j], y = i->second + dy[j];
        if ((0 <= (x) && (x) < (n)) && (0 <= (y) && (y) < (m)) &&
            d[x][y] + 1 == dist) {
          if (ch == s[x][y])
            nx.push_back(make_pair(x, y));
          else if (s[x][y] < ch)
            nx.clear(), nx.push_back(make_pair(x, y)), ch = s[x][y];
        }
      }
    sort((nx).begin(), (nx).end());
    nx.resize(unique((nx).begin(), (nx).end()) - nx.begin());
    cur = nx;
    nx.clear();
    assert(d[cur[0].first][cur[0].second] == dist - 1);
    RES += ch;
  }
}
int main() {
  int k, si = -1, sj = -1, ti = -1, tj = -1;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  for (int i = 0; i < (n); ++i) scanf("%s", s[i]);
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j)
      if (s[i][j] == 'S')
        si = i, sj = j;
      else if (s[i][j] == 'T')
        ti = i, tj = j;
  for (int i = 0; i < (k); ++i) a[25 - i] = 1;
  int it[4] = {26, 26, 26, 26};
  for (int i = 0; i < (4); ++i) {
    int x = ti + dx[i], y = tj + dy[i];
    if ((0 <= (x) && (x) < (n)) && (0 <= (y) && (y) < (m))) {
      if (s[x][y] == 'S') return 0;
      it[i] = s[x][y] - 'a';
    }
  }
  pair<int, string> res = make_pair(INT_MAX, string(10, 'z' + 1));
  do {
    int any = 0;
    for (int i = 0; i < (4); ++i)
      if (a[it[i]]) any = 1, i = 4;
    if (!any) continue;
    f(ti, tj);
    if (d[si][sj] != -1 && d[si][sj] <= res.first) {
      reconstruct(si, sj);
      if (d[si][sj] < res.first ||
          (d[si][sj] == res.first && RES < res.second)) {
        res.first = d[si][sj];
        res.second = RES;
      }
    }
  } while (next_permutation(a, a + 26));
  if (res.first == INT_MAX)
    printf("-1\n");
  else
    printf("%s\n", res.second.c_str());
  return 0;
}
