环上有序尽多购买 D. Berland Fair CodeForces - 1073D

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在Berland年度集市上，Polycarp计划花费有限的货币购买尽可能多的糖果。集市由多个环形排列的摊位组成，每个摊位出售价格不一的糖果。此算法描述了如何计算Polycarp能购买的糖果总数，遵循特定的购买路径和规则。

D. Berland Fair

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

XXI Berland Annual Fair is coming really soon! Traditionally fair consists of nn booths, arranged in a circle. The booths are numbered 11 through nn clockwise with nn being 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

Copy

3 38
5 2 5

output

Copy

input

Copy

5 21
2 4 100 2 6

output

Copy

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.

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mn = 2e5 + 10;
int a[mn], b[mn];
int main()
{	
	int n; ll T;
	scanf("%d %lld", &n, &T);
	
	ll sum = 0;
	int cnt = 0;
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
		if ((sum + a[i]) <= T)
		{
			b[cnt++] = a[i];
			sum += a[i];
		}
	}
	
	if (sum == 0)
	{
		printf("0\n");
		return 0;
	}
	
	ll ans = T / sum * cnt;
	T %= sum;
	
	ll tmp = 0;
	bool flag;
	while (T > 0)
	{
		sum = 0;
		tmp = 0;
		flag = 0;
		for (int i = 0; i < cnt; i++)
		{
			if (sum + b[i] <= T)
			{
				sum += b[i];
				tmp++;
				flag = 1;
			}
		}
		if (!flag)
			break;
		ans = ans + T / sum * tmp;
		T %= sum;
	}
	printf("%lld\n", ans);
	
	return 0;
}