Problem: We have an H \times W grid, where each square is painted white or black in the initial state. Given are strings A_1, A_2, ..., A_H representing the colors of the squares in the initial state. For each pair (i, j) (1 \leq i \leq H, 1 \leq j \leq W), if the j-th character of A_i is `.`, the square at the i-th row and j-th column is painted white; if that character is `#`, that square is painted black.

Among the 2^{HW} ways for each square in the grid to be painted white or black, how many can be obtained from the initial state by performing the operations below any number of times (possibly zero) in any order? Find this count modulo 998,244,353.

* Choose one row, then paint all the squares in that row white.
* Choose one row, then paint all the squares in that row black.
* Choose one column, then paint all the squares in that column white.
* Choose one column, then paint all the squares in that column black.

Constraints

* 1 \leq H, W \leq 10
* |A_i| = W (1 \leq i \leq H)
* All strings A_i consist of `.` and `#`.
* H and W are integers.

Input

Input is given from Standard Input in the following format:


H W
A_1
A_2
\vdots
A_H


Output

Print the answer.

Examples

Input

2 2
#.
.#


Output

15


Input

2 2
.
.#


Output

15


Input

3 3
...
...
...


Output

230


Input

2 4
...
...#


Output

150


Input

6 7
.......
.......
.#.....
..#....
.#.#...
.......


Output

203949910

Solution (in CPP):

Solution:
//Zory-2020
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define mem(x,val) memset(x,val,sizeof x)
#define Swap(a,b,n) for(int I=0;I<=n;I++) swap(a[I],b[I])
#define PC __builtin_popcountll
#ifdef DEBUG
#define DD 1
#else
#define DD 0
#endif
#define Debug if(DD) printf("line %d\n",__LINE__)
#define deb(x) if(DD) cerr<<#x<<'='<<x<<endl
#define debug if(DD) printf
namespace mine
{
	ll qread()
	{
		ll ans=0,f=1;char c=getchar();
		while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
		while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
		return ans*f;
	}
	void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
	void write1(ll num){write(num);putchar(' ');}
	void write2(ll num){write(num);putchar('\n');}
	template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
	template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
	ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
	bool IN(ll x,ll l,ll r){return l<=x and x<=r;}
	void gg(){write2(-1);exit(0);}
	
	const double eps=1e-8;
	const int INF=0x3f3f3f3f;
	const int MOD=998244353;
	int mm(const int x){return x>=MOD?x-MOD:x;}
	template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
	ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
	ll invm(ll x){return qpower(x,MOD-2);}
	const int N=1e2+10;

	ll C[N][N];
	ll f[N][N][2],g[N][N][2];//f=上次删的是行，g=列
	char mp[N][N];
	void main()
	{
		C[0][0]=1;fo(i,1,N-1){ C[i][0]=1;fo(j,1,i) C[i][j]=mm(C[i-1][j-1]+C[i-1][j]); }
		int n=qread(),m=qread();fo(i,0,n-1) scanf("%s",mp[i]);
		fo(mm,0,m) g[n][mm][1]=C[m][mm]*bin(m-mm)%MOD;
		fd(nn,n,0) fd(mm,m,0)
		{
			fo(ad,1,n-nn)
				add(f[nn][mm][0],C[ad+nn][nn]*(g[nn+ad][mm][0]+g[nn+ad][mm][1]*2)%MOD ),
				add(f[nn][mm][1],C[ad+nn][nn]*g[nn+ad][mm][1]%MOD*(bin(ad)-2)%MOD );
			fo(ad,1,m-mm)
				add(g[nn][mm][0],C[ad+mm][mm]*(f[nn][mm+ad][0]+f[nn][mm+ad][1]*2)%MOD ),
				add(g[nn][mm][1],C[ad+mm][mm]*f[nn][mm+ad][1]%MOD*(bin(ad)-2)%MOD );
		}
		ll ans=0;fo(mm,1,m) add(ans,f[0][mm][1]);fo(nn,1,n) add(ans,g[nn][0][1]);
		fo(S1,1,bin(n)-1) fo(S2,1,bin(m)-1)
		{
			bool ok=1;
			fo(i,0,n-1) if(S1&bin(i)){
				bool same=1;int lst=-1;
				fo(j,0,m-1) if(S2&bin(j)) same&=(lst<0 or mp[i][j]==mp[i][lst]),lst=j;
				ok&=(same==0);
			}fo(j,0,m-1) if(S2&bin(j)){
				bool same=1;int lst=-1;
				fo(i,0,n-1) if(S1&bin(i)) same&=(lst<0 or mp[i][j]==mp[lst][j]),lst=i;
				ok&=(same==0);
			}
			if(!ok) continue;
			fo(pp,0,1) add(ans, mm(f[PC(S1)][PC(S2)][pp]+g[PC(S1)][PC(S2)][pp])*invm(C[n][PC(S1)]*C[m][PC(S2)]%MOD)%MOD );
		}write(ans);
	}
};//(ans+MOD)%MOD
signed main()
{
	srand(time(0));
	mine::main();
}