題干：

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In the morning, there are?nn?opportunities to buy shares. The?ii-th of them allows to buy as many shares as you want, each at the price of?sisi?bourles.

In the evening, there are?mm?opportunities to sell shares. The?ii-th of them allows to sell as many shares as you want, each at the price of?bibi?bourles. You can't sell more shares than you have.

It's morning now and you possess?rr?bourles and no shares.

What is the maximum number of bourles you can hold after the evening?

Input

The first line of the input contains three integers?n,m,rn,m,r?(1≤n≤301≤n≤30,?1≤m≤301≤m≤30,?1≤r≤10001≤r≤1000) — the number of ways to buy the shares on the market, the number of ways to sell the shares on the market, and the number of bourles you hold now.

The next line contains?nn?integers?s1,s2,…,sns1,s2,…,sn?(1≤si≤10001≤si≤1000);?sisi?indicates the opportunity to buy shares at the price of?sisi?bourles.

The following line contains?mm?integers?b1,b2,…,bmb1,b2,…,bm?(1≤bi≤10001≤bi≤1000);?bibi?indicates the opportunity to sell shares at the price of?bibi?bourles.

Output

Output a single integer — the maximum number of bourles you can hold after the evening.

Examples

Input

3 4 11 4 2 5 4 4 5 4

Output

26

Input

2 2 50 5 7 4 2

Output

50

Note

In the first example test, you have?1111?bourles in the morning. It's optimal to buy?55shares of a stock at the price of?22?bourles in the morning, and then to sell all of them at the price of?55?bourles in the evening. It's easy to verify that you'll have?2626?bourles after the evening.

In the second example test, it's optimal not to take any action.

解題報告：

簡單的貪心，，剛開始讀錯題了還以為是個背包。。（cfA題出背包？？）寫完發現是個完全背包，所以也是可以貪心的，想了想放到A題還是有些合理的，，再看樣例解釋發現不對，，比想象中的水多了、、、就寫了、、

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int dp[MAX],dpp[MAX],n,m,r,a[MAX],b[MAX]; int main() {cin>>n>>m>>r;for(int i = 1; i<=n; i++) cin>>a[i];for(int i = 1; i<=m; i++) cin>>b[i];int minn = *min_element(a+1,a+n+1);int maxx = *max_element(b+1,b+m+1);int ci = r/minn;printf("%d\n",max(r,r + (maxx-minn)*ci));// for(int i = 0; i<=r; i++) dpp[i] = r;//設置初始狀態，dpp[i]代表有i塊錢時可以擁有的最大錢數，因為啥都不買所以擁有的是r // for(int i = 1; i<=min(n,m); i++) { // for(int j = a[i]; j<=r; j++) { // if(dp[j-a[i]] + b[i]-a[i] > dp[j]) { // dp[j] = dp[j-a[i]] + b[i]-a[i]; // dpp[j] = dpp[j-a[i]] + b[i]-a[i]; // } // } // } // printf("%d\n",dpp[r]);return 0 ; }

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