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
import sys
from array import array  # noqa: F401
from itertools import combinations
from collections import deque


def input():
    return sys.stdin.buffer.readline().decode('utf-8')


n, m, k = map(int, input().split())
chars = (
    ['}' * (m + 2)]
    + ['}' + ''.join('{' if c == 'S' else '|' if c == 'T' else c for c in input().rstrip()) + '}' for _ in range(n)]
    + ['}' * (m + 2)]
)
cbit = [[1 << (ord(c) - 97) for c in chars[i]] for i in range(n + 2)]

si, sj, ti, tj = 0, 0, 0, 0
for i in range(1, n + 1):
    for j in range(1, m + 1):
        if chars[i][j] == '{':
            si, sj = i, j
            cbit[i][j] = 0
        if chars[i][j] == '|':
            ti, tj = i, j


ans = inf = '*' * (n * m)

for comb in combinations([1 << i for i in range(26)], r=k):
    enabled = sum(comb)

    dp = [[inf] * (m + 2) for _ in range(n + 2)]
    dp[ti][tj] = ''
    dq = deque([(ti, tj, '')])
    while dq:
        i, j, s = dq.popleft()
        if dp[i][j] < s:
            continue
        for di, dj in ((i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)):
            if (cbit[di][dj] & enabled) != cbit[di][dj]:
                continue
            pre = chars[di][dj] if cbit[di][dj] else ''
            l = 1 if cbit[di][dj] else 0
            if (len(dp[di][dj]) > len(s) + l or len(dp[di][dj]) == len(s) + l and dp[di][dj] > pre + s):
                dp[di][dj] = pre + s
                if l:
                    dq.append((di, dj, pre + s))

    if len(ans) > len(dp[si][sj]) or len(ans) == len(dp[si][sj]) and ans > dp[si][sj]:
        ans = dp[si][sj]

print(ans if ans != inf else -1)
