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 = 51;
const int maxm = 26;
const int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
struct Position {
  int x, y;
};
Position st, en;
int n, m, k;
char mp[maxn][maxn];
bool hv[maxm], us[maxm];
int hc;
string ph, th;
int ml, tl;
int sd[maxn][maxn], ed[maxn][maxn];
queue<Position> qe;
bool Bound(Position p) { return p.x >= 0 && p.x < n && p.y >= 0 && p.y < m; }
void Bfs(Position s, int dis[maxn][maxn]) {
  int i, j;
  Position nw, nx;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) dis[i][j] = -1;
  }
  dis[s.x][s.y] = 0;
  qe.push(s);
  while (!qe.empty()) {
    nw = qe.front();
    qe.pop();
    for (i = 0; i < 4; i++) {
      nx.x = nw.x + dir[i][0];
      nx.y = nw.y + dir[i][1];
      if (Bound(nx) && dis[nx.x][nx.y] == -1) {
        if (!(mp[nx.x][nx.y] >= 'a' && mp[nx.x][nx.y] <= 'z' &&
              us[mp[nx.x][nx.y] - 'a'] == 0)) {
          qe.push(nx);
          dis[nx.x][nx.y] = dis[nw.x][nw.y] + 1;
        }
      }
    }
  }
}
void FindPath() {
  vector<Position> tmp;
  Position nw, nx;
  char mi = 'z' + 1;
  int step = 1;
  int i;
  qe.push(st);
  while (!qe.empty()) {
    nw = qe.front();
    qe.pop();
    for (i = 0; i < 4; i++) {
      nx.x = nw.x + dir[i][0];
      nx.y = nw.y + dir[i][1];
      if (Bound(nx) && sd[nx.x][nx.y] == step &&
          sd[nx.x][nx.y] + ed[nx.x][nx.y] == tl) {
        sd[nx.x][nx.y] = -1;
        if (mp[nx.x][nx.y] < mi) {
          mi = mp[nx.x][nx.y];
          tmp.clear();
          tmp.push_back(nx);
        } else if (mp[nx.x][nx.y] == mi) {
          tmp.push_back(nx);
        }
      }
    }
    if (qe.empty()) {
      if (tmp.empty()) break;
      while (!tmp.empty()) {
        qe.push(tmp.back());
        tmp.pop_back();
      }
      if (mi != 'T') th += mi;
      mi = 'z' + 1;
      step++;
    }
  }
}
void Run() {
  tl = -1;
  th = "";
  Bfs(st, sd);
  Bfs(en, ed);
  if (sd[en.x][en.y] == -1) return;
  tl = sd[en.x][en.y];
  FindPath();
  if (ml == -1 || ml > tl || (ml == tl && ph > th)) {
    ml = tl;
    ph = th;
  }
}
void Select(int xp, int au) {
  if (au >= k) {
    Run();
    return;
  }
  if (xp >= maxm) {
    return;
  }
  if (hv[xp]) {
    us[xp] = 1;
    Select(xp + 1, au + 1);
    us[xp] = 0;
  }
  Select(xp + 1, au);
}
int main() {
  int i, j;
  while (cin >> n >> m >> k) {
    for (i = 0; i < n; i++) cin >> mp[i];
    memset(hv, 0, sizeof(hv));
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (mp[i][j] >= 'a' && mp[i][j] <= 'z')
          hv[mp[i][j] - 'a'] = 1;
        else if (mp[i][j] == 'S')
          st.x = i, st.y = j;
        else if (mp[i][j] == 'T')
          en.x = i, en.y = j;
      }
    }
    hc = 0;
    for (i = 0; i < maxm; i++)
      if (hv[i]) hc++;
    ml = -1;
    ph = "";
    if (k >= hc) {
      for (i = 0; i < maxm; i++) us[i] = 1;
      Run();
    } else {
      memset(us, 0, sizeof(us));
      Select(0, 0);
    }
    if (ml == -1)
      cout << "-1" << endl;
    else
      cout << ph << endl;
  }
  return 0;
}
