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 = 61;
const int MAXA = 31;
const int INF = 1000000000;
const int MAXM = 10010;
const int w[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
struct P {
  P() {}
  P(int xx, int yy) : x(xx), y(yy) {}
  bool operator==(const P &p) const { return x == p.x && y == p.y; }
  int x, y;
} start, target;
char mp[MAXN][MAXN];
int ch[MAXA];
bool vis[MAXA];
int dist[MAXN][MAXN];
int ctot;
int n, m, k;
int ans;
char path[MAXM], curpath[MAXM];
void init() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; ++i) scanf("%s", mp[i]);
  ctot = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (mp[i][j] == 'S')
        start = P(i, j);
      else if (mp[i][j] == 'T')
        target = P(i, j);
      else {
        if (!vis[mp[i][j] - 'a']) {
          ch[ctot++] = mp[i][j] - 'a';
          vis[mp[i][j] - 'a'] = 1;
        }
      }
  k = min(ctot, k);
}
P que[MAXM];
void bfs() {
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) dist[i][j] = -1;
  que[0] = target;
  dist[target.x][target.y] = 0;
  P p;
  for (int k1 = 0, k2 = 1, k3 = 1, step = 1; k1 < k2; k1 = k3, k3 = k2, ++step)
    for (int i = k1; i < k3; ++i)
      for (int l = 0; l < 4; ++l) {
        p.x = que[i].x + w[l][0];
        p.y = que[i].y + w[l][1];
        if (!(0 <= (p.x) && (p.x) < n && 0 <= (p.y) && (p.y) < m) ||
            dist[p.x][p.y] != -1 ||
            (mp[p.x][p.y] != 'S' && !vis[mp[p.x][p.y] - 'a']))
          continue;
        dist[p.x][p.y] = step;
        que[k2++] = p;
      }
  if (dist[start.x][start.y] == -1 || dist[start.x][start.y] > ans) return;
  que[0] = start;
  int len = 0;
  for (int k1 = 0, k2 = 1, k3 = 1, step = dist[start.x][start.y] - 1; k1 < k2;
       k1 = k3, k3 = k2, --step) {
    char c = 127;
    for (int i = k1; i < k3; ++i)
      for (int l = 0; l < 4; ++l) {
        p.x = que[i].x + w[l][0];
        p.y = que[i].y + w[l][1];
        if (!(0 <= (p.x) && (p.x) < n && 0 <= (p.y) && (p.y) < m) ||
            dist[p.x][p.y] != step)
          continue;
        c = min(c, mp[p.x][p.y]);
      }
    if (c == 'T') break;
    curpath[len++] = c;
    for (int i = k1; i < k3; ++i)
      for (int l = 0; l < 4; ++l) {
        p.x = que[i].x + w[l][0];
        p.y = que[i].y + w[l][1];
        if (!(0 <= (p.x) && (p.x) < n && 0 <= (p.y) && (p.y) < m) ||
            dist[p.x][p.y] != step || mp[p.x][p.y] != c)
          continue;
        dist[p.x][p.y] = -1;
        que[k2++] = p;
      }
  }
  curpath[len] = '\0';
  if (ans > dist[start.x][start.y] || strcmp(curpath, path) < 0) {
    ans = dist[start.x][start.y];
    strcpy(path, curpath);
  }
}
void cal(int step, int pt) {
  if (step == 0) {
    bfs();
    return;
  }
  for (int i = pt; i <= ctot - step; ++i) {
    vis[ch[i]] = 1;
    cal(step - 1, i + 1);
    vis[ch[i]] = 0;
  }
}
void solve() {
  memset(vis, 0, sizeof(vis));
  ans = INF;
  path[0] = 127;
  path[1] = '\0';
  cal(k, 0);
  if (ans == INF)
    puts("-1");
  else
    printf("%s\n", path);
}
int main() {
  init();
  solve();
  return 0;
}
