People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch. 
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins. 

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

題意：有n種硬幣，價值不超過m的商品，每個硬幣對應一個價值w[i]和數量c[i]，求這些硬幣恰好能夠組合成不超過m的價值的方案數。

思路：樓教主的男人八題之一，算是一個經典的問題，定義一個sum陣列。每次填dp[j]時直接由dp[j-w[i]]推出，前提是sum[j-w[i]]<c[i]。 sum每填一行都要清零，sum[j]表示當前物品填充j大小的價值需要至少使用多少價值為w[i]的硬幣數量。

#include <iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int n,m;
int w[105],c[105],num[100005];
bool dp[100005];

int main()
{
    while(cin>>n>>m)
    {
        if(n==0&&m==0)
            break;
        for(int i=0; i<n; i++)
        {
            cin>>w[i];
        }
        for(int i=0; i<n; i++)
        {
            cin>>c[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        int ans=0;
        for(int i=0; i<n; i++)
        {
            memset(num,0,sizeof(num));
            for(int j=w[i]; j<=m; j++)
            {
                if(!dp[j]&&dp[j-w[i]]&&num[j-w[i]]<c[i])//價值為j的還沒湊齊，但是j-w[i]的已經湊齊，並且剩下需要湊齊的數量要小於已有的c[i].
                {
                    dp[j]=1;
                    num[j]=num[j-w[i]]+1;//數量加一。
                    ans++;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}