Problem: A mouse encountered a nice big cake and decided to take a walk across it, eating the berries on top of the cake on its way. The cake is rectangular, neatly divided into squares; some of the squares have a berry in them, and some don't.

The mouse is in a bit of a hurry, though, so once she enters the cake from its northwest corner (the top left cell in the input data), she will only go east (right) or south (down), until she reaches the southeast corner (the bottom right cell). She will eat every berry in the squares she passes through, but not in the other squares.

The mouse tries to choose her path so as to maximize the number of berries consumed. However, her haste and hunger might be clouding her judgement, leading her to suboptimal decisions...

Input

The first line of input contains two integers H and W (1 ≤ H, W ≤ 5), separated by a space, — the height and the width of the cake.

The next H lines contain a string of W characters each, representing the squares of the cake in that row: '.' represents an empty square, and '*' represents a square with a berry.

Output

Output the number of berries the mouse will eat following her strategy.

Examples

Input


4 3
*..
.*.
..*
...


Output


3


Input


4 4
.*..
*...
...*
..*.


Output


2


Input


3 4
..**
*...
....


Output


1


Input


5 5
..*..
.....
**...
**...
**...


Output


1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL)
#define ll long long
#define INF (ll)1e9 + 7
#define INF2 (ll)1e18 + 9
#define MOD (ll)998244353
#define ull unsigned long long
#define PI 3.141592653589793238
#define pb push_back
#define max3(x, y, z) max(x, max(y, z))
#define min3(x, y, z) min(x, min(y, z))
#define mp make_pair
#define mt make_tuple
#define pre(a,b) cout << fixed << setprecision(a) << b  //a for precision b for answer
#define all(x) (x).begin(), (x).end()
#define lb lower_bound //lower_bound(first_iterator, last_iterator, x) – returns an iterator pointing to the first element in the range [first,last) which has a value not less than ‘x’.
#define ub upper_bound //upper_bound(first_iterator, last_iterator, x) – returns an iterator pointing to the first element in the range [first,last) which has a value greater than ‘x’.
#define maxe *max_element
#define mine *min_element
#define acc(x) accumulate(x.begin(), x.end(), 0)
#define dst distance
#define sz(x) (int)x.size()
#define ff first
#define ss second

char grid[10][10];
int h, w;

int main()
{
    IOS;

    cin >> h >> w;
    for (int i = 1; i <= h; i++)
        for (int j = 1; j <= w; j++)
            cin >> grid[i][j];

    int curx = 1, cury = 1, step = 1, ans = 0;
    while (curx <= h and cury <= w and step <= h + w)
    {
        for (int i = 0; i <= step; i++)
        {
            if (grid[curx + i][cury + step - i] == '*' and curx + i <= h and cury + step - i <= w)
            {
                // cout << curx + i << " " << cury + step - i << '\n';
                curx += i, cury += step - i, step = 0, ans++;
                break;
            }
        }
        step++;
    }
    
    if (grid[1][1] == '*') ans++;

    cout << ans << '\n';

    return 0;
}