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 = 100005;
int n, m, k;
char str[55][55];
int v[55][55];
set<long long> se;
int cnt[maxn];
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
string ans;
int sx, sy, ex, ey;
struct node {
  int x, y;
  int s;
  string str;
  node(int _x = 0, int _y = 0, int _s = 0, string _str = "")
      : x(_x), y(_y), s(_s), str(_str) {}
};
bool operator<(const node &a, const node &b) {
  int lena = a.str.length(), lenb = b.str.length();
  if (lena != lenb) return lena > lenb;
  return a.str > b.str;
}
int cot(int n) {
  int ret = 0;
  for (; n; n -= n & (-n)) ret++;
  return ret;
}
void BFS(int sx, int sy, int ex, int ey) {
  priority_queue<node> q;
  q.push(node(sx, sy, 0, ""));
  se.insert(sx * m + sy);
  while (!q.empty()) {
    node temp = q.top();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int tx = temp.x + dir[i][0];
      int ty = temp.y + dir[i][1];
      if (tx == ex && ty == ey) {
        cout << temp.str << endl;
        return;
      }
      if (tx == sx && ty == sy) continue;
      if (tx < 0 || ty < 0 || tx >= n || ty >= m) continue;
      int news = temp.s | (1 << (str[tx][ty] - 'a'));
      if (cot(news) > k) continue;
      string ss = temp.str + str[tx][ty];
      long long tv = news * n * m + tx * m + ty;
      if (se.find(tv) == se.end()) {
        if (v[tx][ty] && ((v[tx][ty] & news) == v[tx][ty])) continue;
        v[tx][ty] = news;
        se.insert(tv);
        q.push(node(tx, ty, news, temp.str + str[tx][ty]));
      }
    }
  }
  cout << "-1" << endl;
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", str[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (str[i][j] == 'S') sx = i, sy = j;
      if (str[i][j] == 'T') ex = i, ey = j;
    }
  }
  BFS(sx, sy, ex, ey);
  return 0;
}
