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;
template <typename T>
int size(T& a) {
  return (int)a.size();
}
const int MAXN = 51;
const int INF = 1000000000;
char s[MAXN][MAXN];
bool can[26];
int n, m, k;
int sx, sy, fx, fy;
int d[MAXN][MAXN];
bool used[MAXN][MAXN];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
bool valid(int y, int x) {
  return x >= 0 && y >= 0 && x < m && y < n &&
         (s[y][x] == 'T' || s[y][x] == 'S' || can[s[y][x] - 'a']) &&
         d[y][x] == INF;
}
struct Qu {
  int qx[MAXN * MAXN];
  int qy[MAXN * MAXN];
  int head, tail;
  Qu() {
    head = 0;
    tail = 0;
  }
  pair<int, int> pop() {
    int x = qx[tail];
    int y = qy[tail];
    ++tail;
    return make_pair(y, x);
  }
  void clear() {
    head = 0;
    tail = 0;
  }
  void operator=(const Qu& que) {
    head = que.head;
    for (int i = (0); i < (head); ++i) qx[i] = que.qx[i], qy[i] = que.qy[i];
  }
  void push(int x, int y) {
    qx[head] = x;
    qy[head] = y;
    head++;
  }
  int size() { return head - tail; }
};
Qu cur, nxt, Q;
int ans = INF;
string res = "";
void brute(int pos, int take) {
  if (pos == 26) {
    Q.clear();
    for (int i = (0); i < (n); ++i)
      for (int j = (0); j < (m); ++j) d[i][j] = INF;
    if (can['y' - 'a']) {
      int br = 1;
    }
    d[fy][fx] = 0;
    Q.push(fx, fy);
    while (size(Q) > 0) {
      pair<int, int> p = Q.pop();
      for (int u = (0); u < (4); ++u)
        if (valid(p.first + dy[u], p.second + dx[u])) {
          Q.push(p.second + dx[u], p.first + dy[u]);
          d[p.first + dy[u]][p.second + dx[u]] = d[p.first][p.second] + 1;
        }
    }
    if (d[sy][sx] == INF) return;
    string path = "";
    cur.push(sx, sy);
    memset((used), (false), sizeof(used));
    while (size(cur) > 0) {
      int minSymbol = INF;
      for (int i = (0); i < (size(cur)); ++i) {
        int x = cur.qx[i];
        int y = cur.qy[i];
        for (int u = (0); u < (4); ++u) {
          int nx = x + dx[u];
          int ny = y + dy[u];
          if (nx >= 0 && ny >= 0 && nx < m && ny < n &&
              d[ny][nx] == d[y][x] - 1 &&
              (s[ny][nx] == 'S' || s[ny][nx] == 'T' || can[s[ny][nx] - 'a'])) {
            minSymbol = min(minSymbol, (int)s[ny][nx]);
          }
        }
      }
      nxt.clear();
      if (minSymbol != 'T') path += (char)(minSymbol);
      for (int i = (0); i < (size(cur)); ++i) {
        int x = cur.qx[i];
        int y = cur.qy[i];
        for (int u = (0); u < (4); ++u) {
          int nx = x + dx[u];
          int ny = y + dy[u];
          if (nx >= 0 && ny >= 0 && nx < m && ny < n &&
              d[ny][nx] == d[y][x] - 1 &&
              (s[ny][nx] == 'S' || s[ny][nx] == 'T' || can[s[ny][nx] - 'a']) &&
              s[ny][nx] == minSymbol && !used[ny][nx]) {
            nxt.push(nx, ny);
            used[ny][nx] = true;
          }
        }
      }
      cur = nxt;
    }
    if (ans > d[sy][sx] || (ans == d[sy][sx] && res > path)) {
      ans = d[sy][sx];
      res = path;
    }
  } else {
    if (take > 0) {
      can[pos] = true;
      brute(pos + 1, take - 1);
    }
    can[pos] = false;
    brute(pos + 1, take);
  }
}
void solve() {
  scanf("%d%d%d\n", &n, &m, &k);
  for (int i = (0); i < (n); ++i) {
    gets(s[i]);
    for (int j = (0); j < (m); ++j)
      if (s[i][j] == 'S')
        sx = j, sy = i;
      else if (s[i][j] == 'T')
        fx = j, fy = i;
  }
  brute(0, k);
  if (ans == INF)
    cout << "-1";
  else
    printf("%s", res.c_str());
}
int main() { solve(); }
