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;
int lx, ly, lk;
char maze[55][55];
map<pair<int, int>, bool> vis;
int sx, sy;
int ex, ey;
inline int ABS(int x) { return x < 0 ? -x : x; }
inline int h(int x, int y) { return ABS(ex - x) + ABS(ey - y); }
struct S {
  int x, y, t, m;
  char path[100];
  S() {}
  S(int _x, int _y, int _t, int _m, char tpath[]) : x(_x), y(_y), t(_t), m(_m) {
    strcpy(path, tpath);
  }
  friend bool operator<(const S& a, const S& b) {
    int ha = h(a.x, a.y), hb = h(b.x, b.y);
    if (ha + a.t != hb + b.t) return ha + a.t > hb + b.t;
    return strcmp(a.path, b.path) > 0;
  }
};
int dir[4][2] = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
char path[100];
char tpath[100];
inline int CB(int x) {
  int cnt = 0;
  while (x) {
    x &= (x - 1);
    cnt++;
  }
  return cnt;
}
void bfs() {
  vis.clear();
  priority_queue<S> que;
  tpath[0] = 0;
  que.push(S(sx, sy, 0, 0, tpath));
  while (!que.empty()) {
    S u = que.top();
    que.pop();
    if (vis[make_pair(u.x * 55 + u.y, u.m)]) continue;
    vis[make_pair(u.x * 55 + u.y, u.m)] = true;
    for (int i = 0; i < 4; i++) {
      int vx = u.x + dir[i][0];
      int vy = u.y + dir[i][1];
      if (vx < 0 || vx >= lx) continue;
      if (vy < 0 || vy >= ly) continue;
      if (maze[vx][vy] == 'S') continue;
      if (maze[vx][vy] == 'T') {
        puts(u.path);
        return;
      }
      int mask = u.m | 1 << (maze[vx][vy] - 'a');
      if (CB(mask) > lk) continue;
      strcpy(tpath, u.path);
      int len = strlen(tpath);
      tpath[len] = maze[vx][vy];
      tpath[len + 1] = 0;
      que.push(S(vx, vy, u.t + 1, mask, tpath));
    }
  }
  puts("-1");
}
int main() {
  while (~scanf("%d%d%d", &lx, &ly, &lk)) {
    for (int i = 0; i < lx; i++) {
      scanf("%s", maze[i]);
      for (int j = 0; j < ly; j++) {
        if (maze[i][j] == 'S') {
          sx = i;
          sy = j;
        } else if (maze[i][j] == 'T') {
          ex = i;
          ey = j;
        }
      }
    }
    bfs();
  }
}
