B. Plus from Picture

You have a given picture with size w×hw×h. Determine if the given picture has a single "+" shape or not. A "+" shape is described below:

• A "+" shape has one center nonempty cell.
• There should be some (at least one) consecutive non-empty cells in each direction (left, right, up, down) from the center. In other words, there should be a ray in each direction.
• All other cells are empty.

Find out if the given picture has single "+" shape.

Input

The first line contains two integers hh and ww (1≤h1≤h, w≤500w≤500) — the height and width of the picture.

The ii-th of the next hh lines contains string sisi of length ww consisting "." and "*" where "." denotes the empty space and "*" denotes the non-empty space.

Output

If the given picture satisfies all conditions, print "YES". Otherwise, print "NO".

You can output each letter in any case (upper or lower).

Examples

input

Copy

5 6
......
..*...
.****.
..*...
..*...

output

Copy

YES

input

Copy

3 5
..*..
****.
.*...

output

Copy

NO

input

Copy

7 7
.......
...*...
..****.
...*...
...*...
.......
.*.....

output

Copy

NO

input

Copy

5 6
..**..
..**..
******
..**..
..**..

output

Copy

NO

input

Copy

3 7
.*...*.
***.***
.*...*.

output

Copy

NO

input

Copy

5 10
..........
..*.......
.*.******.
..*.......
..........

output

Copy

NO

Note

In the first example, the given picture contains one "+".

In the second example, two vertical branches are located in a different column.

In the third example, there is a dot outside of the shape.

In the fourth example, the width of the two vertical branches is 22.

In the fifth example, there are two shapes.

In the sixth example, there is an empty space inside of the shape.

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 3e3 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}
int h, w, coll, colr, rowu, rowd;
bool vis[505][505];
int main()
{
    cin >> h >> w;  
    vector<string> a(h);
    forn(i, h)
        cin >> a[i];
    int fg = 0;
    forn(i, h)
    {
        forn(j, w)
        {
            if (a[i][j] == '*' && i>0 && i<h-1 && a[i+1][j]=='*' && a[i-1][j]=='*' && j>0 && j<w-1 && a[i][j-1]=='*' && a[i][j+1]=='*')
            {
                for (int col1 = j; col1 < w && a[i][col1] != '.'; col1++)
                    vis[i][col1] = 1;
                for (int col2 = j; col2 >= 0 && a[i][col2] != '.'; col2--)
                    vis[i][col2] = 1;
                for (int row1 = i; row1 < h && a[row1][j] != '.'; row1++)
                    vis[row1][j] = 1;
                for (int row2 = i; row2 >= 0 && a[row2][j] != '.'; row2--)
                    vis[row2][j] = 1;
                fg = 1;
                break;
            }
        }
        if (fg)
            break;
    }
    forn(i, h)
    {
        forn(j, w)
        {
            if (a[i][j] == '*' && vis[i][j] == 0)
                fg = 2;
        }
    }
    if (fg == 0 || fg == 2)
        cout << "NO" << endl;
    else
        cout << "YES" << endl;
    return 0;
}