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(300000)

H, W = map(int, input().split())
S = [input() for y in range(H)]
visited = [[False] * W for y in range(H)]

def color(y, x):
    if y < 0 or H <= y or x < 0 or W <= x or visited[y][x]:
        return "@"
    else:
        return S[y][x]


def dfs(y, x):
    b = 0
    w = 0
    nxt = "n"
    if color(y, x) == ".":
        w = 1
        nxt = "#"
    elif color(y, x) == "#":
        b = 1
        nxt = "."
    
    visited[y][x] = True
    if color(y - 1, x) == nxt:
        _b, _w = dfs(y - 1, x)
        b += _b
        w += _w
    if color(y + 1, x) == nxt:
        _b, _w = dfs(y + 1, x)
        b += _b
        w += _w
    if color(y, x - 1) == nxt:
        _b, _w = dfs(y, x - 1)
        b += _b
        w += _w
    if color(y, x + 1) == nxt:
        _b, _w = dfs(y, x + 1)
        b += _b
        w += _w

    return (b, w)

    
ans = 0
for y in range(H):
    for x in range(W):
        b, w = dfs(y, x)
        ans += b * w

print(ans)