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Description

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

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

84

/**poj 1742  多重背包的可行性问题题目大意：给定n种面值的硬币面值分别为wi个数为ci，问用这些硬币可以组成1~m之间的多少面值解题思路：楼教主的男人八题之一，算是一个经典的问题，定义一个sum数组。每次填dp[j]时直接由dp[j-weight[i]]推出，          前提是sum[j-w[i]]
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;int n,m;int w[105],c[105],sum[100005],dp[100005];int main(){    while(~scanf("%d%d",&n,&m)) {        if(n==0&&m==0)break;        for(int i=0;i
      
     
    
   

#include 
   
    using namespace std;const int N = 10 + 100;const int M = 10 + 1e5; int n, m, A[N], C[N];int main() {    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);    while (cin >> n >> m && n && m) {        for (int i = 1; i <= n; i++) cin >> A[i];        for (int i = 1; i <= n; i++) cin >> C[i];        bitset
    
      dp;        dp[0] = 1;        for (int i = 1; i <= n; i++) {                    int k = 1, t = C[i];            while (k < t) {                dp |= (dp << A[i] * k);   // 利用 bitset 的特性取得所有组合和                t  -= k;                k  *= 2;            }            if (t) dp |= (dp << A[i] * t);        }                int ans = 0;        for (int i = 1; i <= m; i++) ans += dp[i];        cout << ans << endl;    }    return 0;}/**    状态定义：  设 f(i,j) 表示是否可使用前 i 种硬币拼成 j    转移方程：  f(i,j) = f(i-1,j-k*Ai), 0<=k<=Ci    边界条件：  f(0,0) = 1, f(0,j) = 0, f(i,0) = 1**/
    
   

#include
   
    using namespace std;const int M = 1e5 + 100;const int N = 110;int  a[N], c[N], book[M], n, m;bool dp[M];//贪心void Coin() {	memset(dp, false, sizeof dp);	for (int i = 1; i <= n; ++i)cin >> a[i];	for (int i = 1; i <= n; ++i)cin >> c[i];	dp[0] = 1;	for (int i = 1; i <= n; ++i) {		memset(book, 0, sizeof book);		for (int j = a[i]; j <= m; ++j)//贪心			if (!dp[j] && dp[j - a[i]] && book[j - a[i]] < c[i])				dp[j] = true, book[j] = book[j - a[i]] + 1;	}	int ans = 0;	for (int i = 1; i <= m; ++i)if (dp[i])++ans;	cout << ans << endl;}int main() {	freopen("in.txt", "r", stdin);	ios::sync_with_stdio(false), cin.tie(0);	while (cin >> n >> m && n) Coin();}