There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

* There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

* 1 \leq H, W \leq 400
* |S_i| = W (1 \leq i \leq H)
* For each i (1 \leq i \leq H), the string S_i consists of characters `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
S_2
:
S_H


Output

Print the answer.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
import sys
sys.setrecursionlimit(500*500)

h, w = map(int, input().split())
grid = []
seen = [[False for y in range(w)] for x in range(h)]

for x in range(h):
    grid.append(input())



blackSeen = 0
whiteSeen = 0
def countFrom(a, b, lastColor):
    global blackSeen, whiteSeen
    if not (0 <= a < h and 0 <= b < w):
        return
    if seen[a][b]:
        return
    #print(a, b)
    

    thisColor = (1 if grid[a][b] == "." else -1)

    if grid[a][b] == "#" and lastColor != -1:
        blackSeen += 1
    elif grid[a][b] == "." and lastColor != 1:
        whiteSeen += 1
    else:
        return
    seen[a][b] = True

    countFrom(a+1, b, thisColor)
    countFrom(a-1, b, thisColor)
    countFrom(a, b+1, thisColor)
    countFrom(a, b-1, thisColor)

tot = 0
for x in range(h):
    for y in range(w):
        blackSeen = 0
        whiteSeen = 0
        countFrom(x, y, 0)
        #print(x, y, blackSeen, whiteSeen)
        tot += blackSeen * whiteSeen

print(tot)

    