CodeForces - 1073D Berland Fair

XXI Berland Annual Fair is coming really soon! Traditionally fair consists of nnbooths, arranged in a circle. The booths are numbered 11 through nn clockwise with nnbeing adjacent to 11. The ii-th booths sells some candies for the price of aiai burles per item. Each booth has an unlimited supply of candies.

Polycarp has decided to spend at most TT burles at the fair. However, he has some plan in mind for his path across the booths:

at first, he visits booth number 11;
if he has enough burles to buy exactly one candy from the current booth, then he buys it immediately;
then he proceeds to the next booth in the clockwise order (regardless of if he bought a candy or not).

Polycarp's money is finite, thus the process will end once he can no longer buy candy at any booth.

Calculate the number of candies Polycarp will buy.

Input

The first line contains two integers nn and TT (1≤n≤2⋅1051≤n≤2⋅105, 1≤T≤10181≤T≤1018) — the number of booths at the fair and the initial amount of burles Polycarp has.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the price of the single candy at booth number ii.

Output

Print a single integer — the total number of candies Polycarp will buy.

Examples

Input

3 38
5 2 5

Output

Input

5 21
2 4 100 2 6

Output

Note

Let's consider the first example. Here are Polycarp's moves until he runs out of money:

Booth 11, buys candy for 55, T=33T=33;
Booth 22, buys candy for 22, T=31T=31;
Booth 33, buys candy for 55, T=26T=26;
Booth 11, buys candy for 55, T=21T=21;
Booth 22, buys candy for 22, T=19T=19;
Booth 33, buys candy for 55, T=14T=14;
Booth 11, buys candy for 55, T=9T=9;
Booth 22, buys candy for 22, T=7T=7;
Booth 33, buys candy for 55, T=2T=2;
Booth 11, buys no candy, not enough money;
Booth 22, buys candy for 22, T=0T=0.

No candy can be bought later. The total number of candies bought is 1010.

In the second example he has 11 burle left at the end of his path, no candy can be bought with this amount.

题目大意：

n种糖果围成一圈，每种糖果每个ai元。初始时你有t元，接着你从1开始疯狂地绕圈。一旦你发现有糖果能买，你就买一个。直到一个糖果都买不起。问最后你买了多少个糖果。

稍微带点技巧的模拟。

若一个周期的和小于剩余的t，就直接买几个周期，不必一个个模拟。

然后遇到买不起的则将他从周期中删除。

注意用long long

#include<cstdio>

#include<queue>

#include<algorithm>

#include<cstring>

#include<cmath>

#include<string>

#include<iostream>

#define lol long long

#define maxn 200000

using namespace std;

lol a[maxn+];

int main()

{

    lol n,t;

    scanf("%lld%lld",&n,&t);

    lol sum=;

    for(int i=;i<=n;i++)

    {

        scanf("%lld",a+i);

        sum+=a[i];

    }

    lol ans=;

    lol cur=n;

    lol num=n;

    while()

    {

        ans+=(t/sum)*num;

        t=t%sum;

        //printf("%lld\n",t);

        for(int i=;;i++)

        {

            //printf("%lld %lld\n",a[(cur+i-1)%n+1],t);

            if(a[(cur+i-)%n+]==-)

                continue;

            if(a[(cur+i-)%n+]>t)

            {

                sum-=a[(cur+i-)%n+];

                num--;

                a[(cur+i-)%n+]=-;

                cur=(cur+i-)%n+;

                break;

            }

            t-=a[(cur+i-)%n+];

            ans++;

        }

        if(num==)

            break;

    }

    printf("%lld\n",ans);

    return ;

}

好久没有更新博客了。

曾经被炒上天的ACM，如今却有些人走茶凉的味道。

正确的事是要坚持的。

巴特西

CodeForces - 1073D Berland Fair

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