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 inf = (1 << 30) - 1;
const long double eps = 1e-9;
const long double pi = fabs(atan2(0.0, -1.0));
int n, m, k;
char a[55][55];
inline void Load() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", a[i]);
}
const int di[4] = {1, 0, 0, -1};
const int dj[4] = {0, 1, -1, 0};
inline bool good(int i, int j) { return 0 <= i && i < n && 0 <= j && j < m; }
inline bool cmp(const pair<int, int> &x, const pair<int, int> &y) {
  return a[x.first][x.second] < a[y.first][y.second];
}
int cnt = 0;
bool can[25000][27];
bool cur[27];
void gen(int i, int j) {
  if (j < 0) return;
  if (i == 26) {
    if (j == 0) {
      for (int k = 0; k < 26; k++) can[cnt][k] = cur[k];
      cnt++;
    }
    return;
  }
  cur[i] = false;
  gen(i + 1, j);
  cur[i] = true;
  gen(i + 1, j - 1);
  cur[i] = false;
}
int d[55][55];
int was[55][55];
pair<int, int> q[500000];
inline void Solve() {
  int si, sj;
  int first, fj;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (a[i][j] == 'S') si = i, sj = j;
      if (a[i][j] == 'T') first = i, fj = j;
    }
  gen(0, k);
  int mdst = 0x7F7F7F7F;
  string ans = "";
  for (int id = 0; id < cnt; id++) {
    int qh = 0, qt = 0;
    memset(d, 0x7F, sizeof(d));
    q[qt++] = make_pair(first, fj);
    d[first][fj] = 0;
    while (qh < qt) {
      int i = q[qh].first;
      int j = q[qh].second;
      qh++;
      for (int l = 0; l < 4; l++) {
        int ni = i + di[l];
        int nj = j + dj[l];
        if (!good(ni, nj)) continue;
        if ('a' <= a[ni][nj] && a[ni][nj] <= 'z' && !can[id][a[ni][nj] - 'a'])
          continue;
        if (d[ni][nj] > d[i][j] + 1) {
          d[ni][nj] = d[i][j] + 1;
          q[qt++] = make_pair(ni, nj);
        }
      }
    }
    if (d[si][sj] > mdst || d[si][sj] == 0x7F7F7F7F) continue;
    string cans = "";
    vector<pair<int, int> > vert;
    vector<pair<int, int> > nvert;
    vert.push_back(make_pair(si, sj));
    memset(was, 0, sizeof(was));
    for (int lane = d[si][sj] - 1; lane >= 0; lane--) {
      char c = 'z' + 1;
      for (int l = 0; l < (int)vert.size(); l++) {
        int i = vert[l].first;
        int j = vert[l].second;
        for (int t = 0; t < 4; t++) {
          int ni = i + di[t];
          int nj = j + dj[t];
          if (!good(ni, nj) || d[ni][nj] != lane) continue;
          c = min(c, a[ni][nj]);
        }
      }
      if ('a' <= c && c <= 'z') cans += c;
      nvert.resize(0);
      for (int l = 0; l < (int)vert.size(); l++) {
        int i = vert[l].first;
        int j = vert[l].second;
        for (int t = 0; t < 4; t++) {
          int ni = i + di[t];
          int nj = j + dj[t];
          if (!good(ni, nj) || d[ni][nj] != lane) continue;
          if (a[ni][nj] == c && was[ni][nj] != lane) {
            was[ni][nj] = lane;
            nvert.push_back(make_pair(ni, nj));
          }
        }
      }
      vert.resize(0);
      for (int l = 0; l < (int)nvert.size(); l++) vert.push_back(nvert[l]);
    }
    if (d[si][sj] < mdst || (d[si][sj] == mdst && cans < ans)) {
      mdst = d[si][sj];
      ans = cans;
    }
  }
  if (mdst == 0x7F7F7F7F)
    printf("-1");
  else
    printf("%s", ans.c_str());
}
int main() {
  Load();
  Solve();
  return 0;
}
