Codeforces Round #417 (Div. 2) C. Sagheer and Nubian Market

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 contains n different items numbered from 1 to n. The i-th item has base cost ai Egyptian pounds. If Sagheer buys k items with indices x1, x2, ..., xk, then the cost of item xj is axj + xj·k for 1 ≤ j ≤ k. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor k.

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 and S (1 ≤ n ≤ 105 and 1 ≤ S ≤ 109) — the number of souvenirs in the market and Sagheer's budget.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 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 these k souvenirs.

Examples

input

3 11
2 3 5

output

2 11

input

4 100
1 2 5 6

output

4 54

input

1 7
7

output

0 0

Note

In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. 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 him 8 pounds, so he cannot buy it.

解题思路：

就是花最少的钱买最多的东西，需要注意的就是公式axj + xj·k ，xj代表的是所选择的数字的下标，所以要先用公式算出其价格，再进行排序。然后就是超时的问题，可以写个二分。

第一次看到c题比b题简单。。怕不是序号出反了

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,b[100009],a[100009],rep,S;
bool check(ll k){
    for(int i = 0; i < n; i++) b[i] = a[i] + k * (i + 1);
    rep = 0;
    sort(b, b + n);
    for(int i = 0; i < k; i++) rep += b[i];
    return rep <= S;
}

int main(){
    ll ans = 0;
    cin >> n >> S;
    for(int i = 0; i < n; i++) cin >> a[i];
    int l = 1, r = n;
    while(l <= r){
        int m = (l + r) / 2;
        if(check(m)){
            l = m + 1;
            ans = rep;
        } else r = m - 1;

    }
    cout << (l - 1) << " " << ans;
    return 0;
}