F(x)

For a decimal number x with n digits (AnAn-1An-2 … A2A1), we define its weight as F(x) = An 2n-1 + An-1 2n-2 + … + A2 2 + A1 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 109)

Output

For every case,you should output “Case #t: “ at first, without quotes. The t is the case number starting from 1. Then output the answer.

Sample Input

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3
0 100
1 10
5 100

Sample Output

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2
3
Case #1: 1
Case #2: 2
Case #3: 13

最开始的思路肯定是搜索

首先是对数位的搜索

然后开始思考另外有什么状态是需要往下面传的，那么就只有前面数的和了，因此是二维的，第一维度是数位，第二个维度是sum;

然后思考记忆化,最直观的就是第二维度就是sum,但是这样子的话每一次都需要根据all进行修改;比较特别的是dp里面的sum记录的是all-sum的值,这样做可以有效的利用memset优化;

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#include <iostream>
#include <cstring>

int dp[12][5005];//dp[pos][sum]:对于位置pos的数位,这里的sum存储的是all-当前pos前缀和的差;
int cur[12];
int all;

using namespace std;
int f(int a){//计算F(a)的值
	if(a==0)return 0;
	int pos=0;
	int able[11];
	while(a){
		able[pos++]=a%10;
		a/=10;
	}
	int result=0;
//	int now=1;
	for(int i=pos-1;i>=0;--i){
		result+=(able[i]<<i);
		//now*=2;
	}
	//cout<<"result:"<<result<<endl;
	return result;
}
// pos:当前的数位,sum 当前的和

/*
	if(pos<0)return 1;合法;
	



*/
int dfs(int pos,int sum,bool limit){//这里的sum表示的是到目前数位的前缀和

	if(pos<0)return sum<=all;
	if(sum>all)return 0;
	
	if(!limit&&dp[pos][all-sum]!=-1)return dp[pos][all-sum];
	
	int up=limit?cur[pos]:9;
	int ans=0;
	for(int i=0;i<=up;++i){
		
		ans+=dfs(pos-1,sum+i*(1<<pos),limit&&cur[pos]==i);
	}
	if(!limit)dp[pos][all-sum]=ans;
//	cout<<pos<<" "<<sum<<" "<<limit<<" "<<ans<<endl;
	return ans;

	
}




int solve(int a,int b){
//	memset(cur,0,sizeof(cur));
 	all=f(a);
	int pos=0;
	if(b==0)return 0;
	while(b){
		cur[pos++]=b%10;
		b/=10;
	}
	// for(int i=pos-1;i>=0;--i){
	// 	cout<<cur[i]<<endl;
	// }
	int result=dfs(pos-1,0,true);


	return result;
}



int main(){
	int t;
	//cin>>t;
	scanf("%d",&t);
	memset(dp,-1,sizeof(dp));
	for(int i=0;i<t;++i){
		int a,b;
		scanf("%d%d",&a,&b);
		//cin>>a>>b;
		int result=solve(a,b);
	//	printf("Case #%d: %d",i+1,result);
	printf("Case #%d: %d\n",i+1,result);
	//	cout<<"Case #"<<i+1<<": "<<result<<endl;
	}
	return 0;


}