Given n integers and a knapsack of weight W,you have to count the number of combinations for which you can add the items inthe knapsack without overflowing the weight.

Input

Input starts with an integer T (≤ 100),denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤30) and W (1 ≤ W ≤ 2 * 10⁹) and the next linewill containn integers separated by spaces. The integers will be nonnegative and less than10⁹.

Output

For each set of input, print the case number and the numberof possible combinations.

Sample input

3

1 1

1

1 1

2

3 10

1 2 4

Sample Output

Case 1: 2

Case 2: 1

Case 3: 8

有ｎ（<=３０）个整数和一个数Ｗ（<=2e9）。问每个数之多用一次并且总和不超过Ｗ的情况下有多少种组合数。

所有的情况有2^30种，必然ＴＬＥ。

将ｎ个数分成两份，在每一份里面统计用这些数能组成多少种整数。然后两种一综合就是答案。

ll a[maxn];
ll b[maxn];
ll t[maxn];
int main(){
	int T;

	cin>>T;
	for(int tt=1;tt<=T;tt++){
		memset(t,0,sizeof(t));
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int n;
		ll w;
		scanf("%d%lld",&n,&w);
		int ta=n>>1;
		int tb=n-ta;
		int l=1<<ta;
		int r=1<<tb;
		for(int i=0;i<n;i++)
			scanf("%lld",&t[i]);

		for(int i=0;i<l;i++){
			for(int j=0;j<ta;j++){
				if(i&(1<<j))
					a[i]+=t[j];
			}
		}
		for(int i=0;i<r;i++){
			for(int j=0;j<tb;j++){
				if(i&(1<<j))
					b[i]+=t[ta+j];
			}
		}
		ll ans=0;
		sort(b,b+r);
		for(int i=0;i<l;i++){
			if(w-a[i]>=0){
				ans+=upper_bound(b,b+r,w-a[i])-b;
			}
		}
		printf("Case %d: %lld\n",tt,ans);
	}
	return 0;
}