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 SX, SY, TX, TY;
int N, M, K;
char V[52][52];
vector<vector<int> > C;
vector<int> cur;
void gen(int sv, int how) {
  if (sv >= 26 && cur.size() == K) C.push_back(cur);
  if (sv >= 26 || how < 0) return;
  gen(sv + 1, how);
  if (how > 0) {
    cur.push_back(sv);
    gen(sv + 1, how - 1);
    cur.pop_back();
  }
}
int bestl = (int)1E9;
string best;
bool can[300];
int LEV[50][50];
deque<int> QX, QY;
string dp[50][50];
string wrd;
vector<int> PX, PY;
vector<int> _PX, _PY;
int u1[50][50], u1s = 0;
string dfs(int x, int y) {
  PX.clear();
  PY.clear();
  wrd = "";
  PX.push_back(x);
  PY.push_back(y);
  while (1) {
    ++u1s;
    if (PX[0] == TX && PY[0] == TY) return wrd;
    _PX.clear();
    _PY.clear();
    char USE = 'z' + 1;
    for (int i = 0; i < PX.size(); ++i) {
      x = PX[i];
      y = PY[i];
      if (x - 1 >= 0 && can[V[x - 1][y]] && LEV[x - 1][y] == LEV[x][y] - 1)
        USE = min(USE, V[x - 1][y]);
      if (y - 1 >= 0 && can[V[x][y - 1]] && LEV[x][y - 1] == LEV[x][y] - 1)
        USE = min(USE, V[x][y - 1]);
      if (x + 1 < N && can[V[x + 1][y]] && LEV[x + 1][y] == LEV[x][y] - 1)
        USE = min(USE, V[x + 1][y]);
      if (y + 1 < M && can[V[x][y + 1]] && LEV[x][y + 1] == LEV[x][y] - 1)
        USE = min(USE, V[x][y + 1]);
    }
    wrd += USE;
    for (int i = 0; i < PX.size(); ++i) {
      x = PX[i];
      y = PY[i];
      if (x - 1 >= 0 && can[V[x - 1][y]] && LEV[x - 1][y] == LEV[x][y] - 1 &&
          USE == V[x - 1][y]) {
        if (u1[x - 1][y] != u1s) {
          _PX.push_back(x - 1);
          _PY.push_back(y);
          u1[x - 1][y] = u1s;
        }
      }
      if (y - 1 >= 0 && can[V[x][y - 1]] && LEV[x][y - 1] == LEV[x][y] - 1 &&
          USE == V[x][y - 1]) {
        if (u1[x][y - 1] != u1s) {
          _PX.push_back(x);
          _PY.push_back(y - 1);
          u1[x][y - 1] = u1s;
        }
      }
      if (x + 1 < N && can[V[x + 1][y]] && LEV[x + 1][y] == LEV[x][y] - 1 &&
          USE == V[x + 1][y]) {
        if (u1[x + 1][y] != u1s) {
          _PX.push_back(x + 1);
          _PY.push_back(y);
          u1[x + 1][y] = u1s;
        }
      }
      if (y + 1 < M && can[V[x][y + 1]] && LEV[x][y + 1] == LEV[x][y] - 1 &&
          USE == V[x][y + 1]) {
        if (u1[x][y + 1] != u1s) {
          _PX.push_back(x);
          _PY.push_back(y + 1);
          u1[x][y + 1] = u1s;
        }
      }
    }
    PX = _PX;
    PY = _PY;
  }
}
string BFS(vector<int> U) {
  int i, j;
  for (i = 0; i < N; ++i)
    for (j = 0; j < M; ++j) dp[i][j].clear();
  memset(LEV, -1, sizeof LEV);
  QX.clear();
  QY.clear();
  for (i = 0; i < U.size(); ++i) can[U[i] + 'a'] = true;
  can['S'] = can['T'] = true;
  LEV[TX][TY] = 1;
  QX.push_back(TX);
  QY.push_back(TY);
  while (!QX.empty()) {
    int x = QX[0], y = QY[0];
    QX.pop_front();
    QY.pop_front();
    if (x - 1 >= 0 && can[V[x - 1][y]] && LEV[x - 1][y] == -1) {
      LEV[x - 1][y] = LEV[x][y] + 1;
      QX.push_back(x - 1);
      QY.push_back(y);
    }
    if (y - 1 >= 0 && can[V[x][y - 1]] && LEV[x][y - 1] == -1) {
      LEV[x][y - 1] = LEV[x][y] + 1;
      QX.push_back(x);
      QY.push_back(y - 1);
    }
    if (x + 1 < N && can[V[x + 1][y]] && LEV[x + 1][y] == -1) {
      LEV[x + 1][y] = LEV[x][y] + 1;
      QX.push_back(x + 1);
      QY.push_back(y);
    }
    if (y + 1 < M && can[V[x][y + 1]] && LEV[x][y + 1] == -1) {
      LEV[x][y + 1] = LEV[x][y] + 1;
      QX.push_back(x);
      QY.push_back(y + 1);
    }
    if (LEV[SX][SY] != -1) break;
  }
  string ret = "NO";
  if (LEV[SX][SY] - LEV[TX][TY] - 1 <= bestl)
    if (LEV[SX][SY] != -1) {
      ret = dfs(SX, SY);
      ret = ret.substr(0, (int)ret.size() - 1);
    }
  for (i = 0; i < U.size(); ++i) can[U[i] + 'a'] = false;
  can['S'] = can['T'] = false;
  return ret;
}
bool IN[300];
int main() {
  int i, j, k, l, B = 0;
  scanf("%d%d%d", &N, &M, &K);
  for (i = 0; i < N; ++i) scanf("%s", &V[i]);
  for (i = 0; i < N; ++i)
    for (j = 0; j < M; ++j) {
      if (V[i][j] == 'S') {
        SX = i;
        SY = j;
      } else if (V[i][j] == 'T') {
        TX = i;
        TY = j;
      } else {
        if (!IN[V[i][j]]) ++B;
        IN[V[i][j]] = 1;
      }
    }
  K = min(K, B);
  gen(0, K);
  for (i = 0; i < C.size(); ++i) {
    bool OK = true;
    for (j = 0; j < C[i].size(); ++j)
      if (!IN['a' + C[i][j]]) {
        OK = false;
        break;
      }
    if (!OK) continue;
    string T = BFS(C[i]);
    if (T == "NO") continue;
    if (T.size() < bestl || (T.size() == bestl && T < best)) {
      best = T;
      bestl = best.size();
    }
  }
  if (bestl > 1000000)
    cout << "-1\n";
  else
    cout << best << '\n';
  return 0;
}
