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;
struct Node {
  int r, c;
};
struct Node2 {
  int r, c, pr, pc;
};
int N, M, K, SR, SC, TR, TC;
char grid[52][52], cost2[50 * 50 + 2];
int cost[52][52], SP, mini;
bool vis[52][52], available[512];
int dr[] = {-1, 1, 0, 0}, dc[] = {0, 0, 1, -1};
Node parent[52][52];
string ans;
bool operator<(const Node2 &a, const Node2 &b) {
  if (cost[a.r][a.c] != cost[b.r][b.c]) return cost[a.r][a.c] > cost[b.r][b.c];
  if (grid[a.r][a.c] != grid[b.r][b.c]) return grid[a.r][a.c] < grid[b.r][b.c];
  if (a.r != b.r) return a.r < b.r;
  if (a.c != b.c) return a.c < b.c;
  if (a.pr != b.pr) return a.pr < b.pr;
  return a.pc < b.pc;
}
void getCoordinates(int &R, int &C, char x, char y) {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++)
      if (grid[i][j] == x) R = i, C = j, grid[i][j] = y;
}
void bfs() {
  SP = 1 << 30;
  memset(vis, 0, sizeof vis);
  memset(cost, 63, sizeof cost);
  queue<Node> q;
  q.push(Node{TR, TC});
  cost[TR][TC] = 0;
  while (!q.empty()) {
    Node cur = q.front(), t;
    q.pop();
    if (cur.r == SR && cur.c == SC) {
      SP = cost[SR][SC];
      return;
    }
    if (vis[cur.r][cur.c]) continue;
    vis[cur.r][cur.c] = true;
    for (int i = 0; i < 4; i++) {
      t = Node{cur.r + dr[i], cur.c + dc[i]};
      if (t.r >= 0 && t.r < N && t.c >= 0 && t.c < M && !vis[t.r][t.c] &&
          available[(int)grid[t.r][t.c]] &&
          cost[t.r][t.c] > cost[cur.r][cur.c] + 1) {
        cost[t.r][t.c] = cost[cur.r][cur.c] + 1;
        q.push(t);
      }
    }
  }
}
void dijkstra() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) parent[i][j] = Node{-1, -1};
  if (cost[SR][SC] >= 1 << 20) return;
  memset(vis, 0, sizeof vis);
  memset(cost2, 'z' + 5, sizeof cost2);
  set<Node2> s;
  s.insert(Node2{SR, SC, -1, -1});
  while (!s.empty()) {
    Node2 cur = *s.begin();
    s.erase(s.begin());
    if (vis[cur.r][cur.c] ||
        cost2[SP - cost[cur.r][cur.c]] < grid[cur.r][cur.c])
      continue;
    vis[cur.r][cur.c] = true;
    cost2[SP - cost[cur.r][cur.c]] = grid[cur.r][cur.c];
    parent[cur.r][cur.c] = Node{cur.pr, cur.pc};
    if (cur.r == TR && cur.c == TC) return;
    for (int i = 0; i < 4; i++) {
      Node2 t = Node2{cur.r + dr[i], cur.c + dc[i], cur.r, cur.c};
      if (t.r >= 0 && t.r < N && t.c >= 0 && t.c < M &&
          cost[t.r][t.c] + 1 == cost[cur.r][cur.c])
        s.insert(t);
    }
  }
}
string getPath() {
  Node cur{TR, TC};
  string ret = "";
  while (cur.r != -1) {
    ret += grid[cur.r][cur.c];
    cur = parent[cur.r][cur.c];
  }
  ret = ret.substr(1, (int)ret.size() - 2);
  reverse(ret.begin(), ret.end());
  return ret;
}
void choose(int k, char from) {
  if (k == K) {
    bfs();
    dijkstra();
    string best = getPath();
    if (SP < mini || (SP == mini && best < ans)) mini = SP, ans = best;
    return;
  }
  for (char i = from; i <= 'z'; i++) {
    available[(int)i] = true;
    choose(k + 1, i + 1);
    available[(int)i] = false;
  }
}
int main() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) parent[i][j] = Node{-1, -1};
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 0; i < N; i++) scanf("%s", grid[i]);
  getCoordinates(SR, SC, 'S', 'a' - 1);
  getCoordinates(TR, TC, 'T', 'z' + 1);
  available['a' - 1] = available['z' + 1] = true;
  mini = 1 << 30;
  choose(0, 'a');
  if (mini == 1 << 30)
    printf("-1\n");
  else
    printf("%s\n", ans.c_str());
}
