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;
int bcount[1 << 12];
set<pair<int, int> > done;
string Map[60];
int n, m, k;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
int ex, ey;
int absval(int x) { return x < 0 ? -x : x; }
int dist(int ax, int ay, int bx, int by) {
  return absval(ax - bx) + absval(ay - by);
}
struct pqitem {
  int x, y;
  string S;
  int mask;
  bool operator<(const pqitem &b) const {
    if (S.size() + dist(x, y, ex, ey) > b.S.size() + dist(b.x, b.y, ex, ey))
      return 1;
    if (S.size() + dist(x, y, ex, ey) < b.S.size() + dist(b.x, b.y, ex, ey))
      return 0;
    if (S.size() + dist(x, y, ex, ey) == b.S.size() + dist(b.x, b.y, ex, ey) &&
        S.compare(b.S) > 0)
      return 1;
    else
      return 0;
  }
};
int getbcount(int m) {
  int ret = 0;
  while (m > 0) {
    ret += bcount[m & ((1 << 10) - 1)];
    m >>= 10;
  }
  return ret;
}
int main() {
  for (int i = 1; i <= (1 << 10); i++)
    bcount[i] = bcount[i / 2] + (i & 1 ? 1 : 0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> Map[i];
  int x, y;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (Map[i][j] == 'S') {
        x = i;
        y = j;
      }
      if (Map[i][j] == 'T') {
        ex = i;
        ey = j;
      }
    }
  priority_queue<pqitem> q;
  q.push((pqitem){x, y, "", 0});
  bool found = false;
  while (!q.empty()) {
    pqitem cur = q.top();
    q.pop();
    if (cur.x == ex && cur.y == ey) {
      cout << cur.S << endl;
      found = true;
      break;
    }
    if (done.find(make_pair(cur.x * m + cur.y, cur.mask)) != done.end())
      continue;
    done.insert(make_pair(cur.x * m + cur.y, cur.mask));
    for (int d = 0; d < 4; d++) {
      pqitem nxt = cur;
      nxt.x += dx[d];
      nxt.y += dy[d];
      if (!(nxt.x >= 0 && nxt.x <= n - 1 && nxt.y >= 0 && nxt.y <= m - 1))
        continue;
      if (Map[nxt.x][nxt.y] != 'T') {
        nxt.mask |= (1 << (Map[nxt.x][nxt.y] - 'a'));
        (nxt.S).push_back(Map[nxt.x][nxt.y]);
      }
      if (getbcount(nxt.mask) <= k) {
        q.push(nxt);
      }
    }
  }
  if (!found) printf("-1\n");
}