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
#!/usr/bin/env python
# -*- coding: utf-8 -*-

# AISing Programming Contest 2019：C - Alternating Path
C = '#.'
CR = {'#':0, '.':1}

H, W = [int(i) for i in raw_input().strip().split(' ')]
S = []
for _ in range(H):
    # S.append(raw_input().strip())
    S.append([CR[c] for c in raw_input().strip()])


def find(h, w, sumi):
    while True:
        if sumi[h][w] == False:
            return (h, w)
        w += 1
        if w >= W:
            w = 0
            h += 1
            if h >= H:
                return None
    return None


'''
def chain((h, w), t, c, sumi):
    if 0 <= h < H and 0 <= w < W:
        pass
    else:
        return c
    if sumi[h][w]:
        return c
    if S[h][w] == t:
        sumi[h][w] = True
        c[t] += 1
        for ptr in [(-1, 0), (0, 1), (1, 0), (0, -1)]:
            c = chain((h+ptr[0], w+ptr[1]), 1-t, c, sumi)
        return c
    return c
'''


# dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
dd = [(0, 1), (1, 0), (0, -1), (-1, 0)]

def find_chain((hh, ww), t, sumi):
    # chain_sumi = [[False for _ in range(W)] for _ in range(H)]
    chain_sumi = set()
    h, w = hh, ww
    sumi[h][w] = True
    chain_sumi.add((h, w))
    c = [0, 0]
    c[t] = 1
    d = 0
    stack = []
    while True:
        while True:
            if d >= 4:
                break
            hd = h + dd[d][0]
            wd = w + dd[d][1]
            d += 1
            if 0 <= hd < H and 0 <= wd < W:
                if (hd, wd) not in chain_sumi:
                    td = S[hd][wd]
                    if td == t-((hh+ww+hd+wd)&1):
                        sumi[hd][wd] = True
                        chain_sumi.add((hd, wd))
                        c[td] += 1
                        stack.append([h, w, d])
                        h , w, d = hd, wd, 0
        if len(stack) == 0:
            break
        h, w, d = stack.pop()
    # print 'sumi :', sumi
    # print 'chain_sumi :', chain_sumi
    return c


def main():
    sumi = [[False for _ in range(W)] for _ in range(H)]
    h, w = 0, 0
    n = 0
    while True:
        # print 'sumi:', sumi
        result = find(h, w, sumi)
        if not result:
            break
        h, w = result[0], result[1]
        # print 'found:', h, w
        c = find_chain((h, w), S[h][w], sumi)
        # print c
        n += c[0] * c[1]
    print n


if __name__ == '__main__':
    main()
