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>
int n, m, k;
char A[64][64];
int d[4];
char bestpath[3000];
int bestlen = -1;
int sx = -1, sy = -1, tx = -1, ty = -1;
int kolejka[3000];
int kval[100][100];
int prev[100][100];
int ks, kk;
int main(void) {
  memset(A, 0, sizeof(A));
  scanf("%d %d %d", &n, &m, &k);
  int i, j;
  for (i = 1; i <= n; i++) {
    scanf("%s", &A[i][1]);
    for (j = 1; j <= m; j++) {
      if (A[i][j] == 'S') {
        sy = i;
        sx = j;
      }
      if (A[i][j] == 'T') {
        ty = i;
        tx = j;
      }
    }
  }
  *bestpath = 0;
  for (d[0] = 'a'; d[0] <= 'z'; d[0]++)
    for (d[1] = d[0]; d[1] <= 'z'; d[1]++)
      for (d[2] = d[1]; d[2] <= 'z'; d[2]++)
        for (d[3] = d[2]; d[3] <= 'z'; d[3]++) {
          int diff = 0;
          int used[300];
          memset(used, 0, sizeof(used));
          for (i = 0; i < 4; i++) {
            if (used[d[i]] == 0) {
              diff++;
            }
            used[d[i]] = 1;
          }
          if (diff > k) {
            continue;
          }
          used['T'] = 1;
          memset(kval, 0xff, sizeof(kval));
          ks = kk = 0;
          kval[sy][sx] = 0;
          kolejka[kk++] = sy * 100 + sx;
          int currentlen = -1;
          int inklen = 0;
          int ddy[4] = {0, -1, 0, 1};
          int ddx[4] = {-1, 0, 1, 0};
          while (ks != kk) {
            int p = kolejka[ks];
            int py = p / 100;
            int px = p % 100;
            ks++;
            if (px == tx && py == ty) {
              currentlen = kval[py][px];
              break;
            }
            for (i = 0; i < 4; i++) {
              int nx = px + ddx[i];
              int ny = py + ddy[i];
              if (!used[A[ny][nx]]) {
                continue;
              }
              if (kval[ny][nx] != -1) {
                continue;
              }
              prev[ny][nx] = p;
              kval[ny][nx] = kval[py][px] + 1;
              kolejka[kk++] = ny * 100 + nx;
            }
          }
          if (currentlen != -1) {
          }
          if (currentlen != -1 &&
              (bestlen == -1 || currentlen - 1 <= bestlen)) {
            currentlen--;
            char tpath[3000];
            memset(tpath, '-', sizeof(tpath));
            tpath[currentlen] = 0;
            int ts = 0, te = 0;
            int avail[64][64], good[64][64];
            memset(avail, 0, sizeof(avail));
            memset(good, 0, sizeof(good));
            avail[sy][sx] = 1;
            good[ty][tx] = 1;
            int kolejka2[3000];
            int ks2, ke2;
            ks2 = ke2 = 0;
            kolejka2[ke2++] = ty * 100 + tx;
            good[ty][tx] = 1;
            while (ks2 != ke2) {
              int p = kolejka2[ks2++];
              int py = p / 100;
              int px = p % 100;
              int k;
              for (k = 0; k < 4; k++) {
                int nx = px + ddx[k];
                int ny = py + ddy[k];
                if (kval[ny][nx] == kval[py][px] - 1 && !good[ny][nx]) {
                  good[ny][nx] = 1;
                  kolejka2[ke2++] = ny * 100 + nx;
                }
              }
            }
            ts = 1;
            for (i = 1; i <= currentlen; i++) {
              te = ts;
              while (te < kk &&
                     kval[kolejka[te] / 100][kolejka[te] % 100] == i) {
                te++;
              }
              bool found = false;
              for (j = 'a'; j <= 'z'; j++) {
                if (used[j]) {
                  int s;
                  for (s = ts; s <= te - 1; s++) {
                    int p = kolejka[s];
                    int py = p / 100;
                    int px = p % 100;
                    if (A[py][px] != j) {
                      continue;
                    }
                    if (!good[py][px]) {
                      continue;
                    }
                    int k;
                    for (k = 0; k < 4; k++) {
                      int nx = px + ddx[k];
                      int ny = py + ddy[k];
                      if (kval[ny][nx] == i - 1 && avail[ny][nx]) {
                        avail[py][px] = 1;
                        found = true;
                      }
                    }
                  }
                }
                if (found) {
                  break;
                }
              }
              tpath[i - 1] = j;
              ts = te;
            }
            if (bestlen == -1 || currentlen < bestlen) {
              bestlen = currentlen;
              strcpy(bestpath, tpath);
            } else {
              if (strcmp(tpath, bestpath) < 0) {
                strcpy(bestpath, tpath);
              }
            }
          }
        }
  if (bestlen == -1) {
    printf("-1\n");
  } else {
    printf("%s\n", bestpath);
  }
  return 0;
}
