Sagheer and Nubian Market （二分）

On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It containsn different items numbered from1 to n. The i-th item has base cost a_i Egyptian pounds. If Sagheer buysk items with indicesx₁, x₂, ..., x_k, then the cost of itemx_j isa_xj + x_j·k for1 ≤ j ≤ k. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factork.

Sagheer wants to buy as many souvenirs as possible without paying more than S Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?

Input

The first line contains two integers n andS (1 ≤ n ≤ 10⁵ and1 ≤ S ≤ 10⁹) — the number of souvenirs in the market and Sagheer's budget.

The second line contains n space-separated integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵) — the base costs of the souvenirs.

Output

On a single line, print two integers k,T — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy thesek souvenirs.

Example

Input

3 112 3 5

Output

2 11

Input

4 1001 2 5 6

Output

4 54

Input

1 77

Output

0 0

Note

In the first example, he cannot take the three items because they will cost him[5, 9, 14] with total cost28. If he decides to take only two items, then the costs will be[4, 7, 11]. So he can afford the first and second items.

In the second example, he can buy all items as they will cost him [5, 10, 17, 22].

In the third example, there is only one souvenir in the market which will cost him8 pounds, so he cannot buy it.

题意：

有一个人要去商店买礼品，商店一共有N种商品，他带有S元钱，而每种商品的最终价格为

sum=val（i）+i*k  val( i )为第i件商品的基本价格，i为商品的编号（从1开始），k为这个人要买的礼品数量

尽可能买多个礼品，求礼品数和钱数；

小坑：数据类型用long long

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#define MAX 100005using namespace std;typedef long long ll;ll val[MAX];ll arr[MAX];ll sum;ll n,m;void cost(int num){    memset(arr,0,sizeof(arr));    ll i;    for(i=1;i<=n;i++)        arr[i]=val[i]+i*num;    sort(arr+1,arr+n+1);//按从小到大排序}ll solve(){//二分法枚举要买的礼品数    ll l=0,r=n;    while(l<r){        int mid=(l+r+1)/2;        cost(mid);//计算每种物品的实际花费        ll i,s=0;        for(i=1;i<=mid;i++)//求总花费            s+=arr[i];        if(s>m) r=mid-1;        else l=mid;    }    return l;}int main(){    scanf("%lld%lld",&n,&m);    ll i;    for(i=1;i<=n;i++)        scanf("%lld",&val[i]);    ll k=solve();    cost(k);    printf("%lld ",k);    for(i=1;i<=k;i++)        sum+=arr[i];    printf("%lld\n",sum);return 0;}