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;
struct node {
  int x, y, g, h, t;
  string p;
  bool operator<(const node& obj) const {
    if (g + h != obj.g + obj.h) return g + h > obj.g + obj.h;
    return p > obj.p;
  }
};
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};
priority_queue<node> q;
set<pair<pair<int, int>, int> > v;
char a[51][55];
int main(void) {
  int n, m, s, sx, sy, tx, ty;
  scanf("%d %d %d", &n, &m, &s);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", a[i] + 1);
    for (int j = 1; j <= m; ++j) {
      if (a[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (a[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  node b;
  b.x = sx;
  b.y = sy;
  b.g = 0;
  b.h = abs(sx - tx) + abs(sy - ty);
  b.t = 0;
  b.p = "";
  q.push(b);
  while (!q.empty()) {
    b = q.top();
    if (b.h == 0) break;
    q.pop();
    pair<pair<int, int>, int> temp = make_pair(make_pair(b.x, b.y), b.t);
    if (v.count(temp) > 0) continue;
    v.insert(temp);
    for (int k = 0; k < 4; ++k) {
      int i = b.x + dx[k], j = b.y + dy[k];
      if ((i < 1) || (i > n) || (j < 1) || (j > m) || (a[i][j] == 'S'))
        continue;
      node e = b;
      e.x += dx[k];
      e.y += dy[k];
      ++e.g;
      e.h = abs(i - tx) + abs(j - ty);
      if (a[i][j] == 'T') {
        q.push(e);
        continue;
      }
      e.t |= (1 << (a[i][j] - 'a'));
      int temp = e.t, c = 0;
      while (temp > 0) {
        c += temp % 2;
        temp /= 2;
      }
      if (c > s) continue;
      e.p += a[i][j];
      q.push(e);
    }
  }
  if (!q.empty())
    cout << q.top().p << endl;
  else
    printf("-1\n");
  return 0;
}
