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

Source: codeforces | Difficulty: VERY_HARD
Tags: Greedy algorithms

Graph algorithms

Shortest paths


Solutions:
import sys
from array import array
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)