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;
struct node {
  string route;
  int x, y, hash, alr;
  node(int x_, int y_, int hash_, int alr_, string route_) {
    x = x_;
    y = y_;
    hash = hash_;
    alr = alr_;
    route = route_;
  }
  friend bool operator<(const node &a, const node &b) {
    if (a.route.length() != b.route.length())
      return a.route.length() > b.route.length();
    return a.route > b.route;
  }
};
int n, m, k;
set<int> visit[MAXN][MAXN];
char s[MAXN][MAXN];
int pre[MAXN][MAXN];
priority_queue<node> q;
bool inside(int x, int y) { return (x >= 0 && x < n && y >= 0 && y < m); }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
void bfs(int sx, int sy) {
  q.push(node(sx, sy, 0, 0, ""));
  while (!q.empty()) {
    node u = q.top();
    q.pop();
    int x = u.x, y = u.y;
    for (int i = 0; i < 4; i++) {
      int lx = x + dx[i];
      int ly = y + dy[i];
      if (!inside(lx, ly)) continue;
      if (s[lx][ly] == 'T') {
        cout << u.route << endl;
        exit(0);
      }
      if (s[lx][ly] == 'S') continue;
      int lhash = u.hash | (1 << (s[lx][ly] - 'a'));
      if (visit[lx][ly].find(lhash) != visit[lx][ly].end()) continue;
      if (pre[lx][ly] != 0 && (pre[lx][ly] & lhash) == pre[lx][ly]) continue;
      pre[lx][ly] = lhash;
      int newalr = u.alr + 1 - ((u.hash >> (s[lx][ly] - 'a') & 1));
      if (newalr > k) continue;
      visit[lx][ly].insert(lhash);
      q.push(node(lx, ly, lhash, newalr, u.route + s[lx][ly]));
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &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') bfs(i, j);
  printf("-1\n");
  return 0;
}
