Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways. 

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!

Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

1 2

1 3

1 4

2 2

2 3

2 4

2 11

4 11

0 0

Sample Output

1

0

1

2

3

5

144

51205

 

题意:求一个h*w的正方形分割成若干个1*2的小正方形 有多少种方案。

思路:对于任意一种方案，考虑以行为界限，dp[i][j]表示前i行 当前行状态为二进制j(1表示竖着放的正方形的上半部分，0表示其他情况)的方案数

dp[i][j]+=dp[i-1][k] 

当且仅当 j&k==0 : 当k为1时 保证对应j的0。 j|k 0的个数必须为偶数 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
ll dp[15][1<<11]; //前i行 状态为j的方案数 
int isodd[1<<11]; //判断是否有偶数个0 
void init(int m){ //预处理 
    for(int i=0;i<(1<<m);i++){
        int cnt=0; int f=1;
        for(int j=0;j<m;j++){
            if((i>>j)&1){
                if(cnt&1) f=0;
                cnt=0;
            }else ++cnt;
        }
        if(cnt&1) f=0;
        isodd[i]=f;
    }
}
int main(){
    ios::sync_with_stdio(false);
    int w,h;
    while(cin>>h>>w){
        if(w==0&&h==0) break;
        memset(dp,0,sizeof(dp));
        init(w);
        dp[0][0]=1;
        for(int i=1;i<=h;i++){
            for(int j=0;j<(1<<w);j++){
                for(int k=0;k<(1<<w);k++)
                    if((j&k)==0&&isodd[j|k])
                        dp[i][j]+=dp[i-1][k];
            }
        }
        cout<<dp[h][0]<<endl;
    } 
    return 0;
}