CodeForces812C 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 1to n. The i-th item has base cost a_i Egyptian pounds. If Sagheer buys k items with indices x₁, x₂, ..., x_k, then the cost of item x_j is a_xj + x_j·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, ..., 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 these k souvenirs.

Example

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.

很容易发现可以装走的个数是满足二分的。。。

因为如果k个可以装走，那么k-a个一定可以装走。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <cmath>
#include <cctype>

using namespace std;

const int N = 1e5 + 10;

typedef long long ll;

ll n, s, a[N], b[N], l, r, ret, ans, out;

bool check(ll k) {
    ret = 0;
    for(int i = 1 ; i <= n ; i ++) {
        b[i] = a[i] + i * k;
    }
    sort(b + 1, b + 1 + n);
    for(int i = 1 ; i <= k ; i ++) {
        ret += b[i];
    }
    return ret <= s;
}

int main() {
    ios :: sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> s;
    for(int i = 1 ; i <= n ; i ++) {
        cin >> a[i];
    }
    l = 0, r = n;
    while(l <= r) {
        int m = (l + r) >> 1;
        if(check(m)) {
            ans = m;
            out = ret;
            l = m + 1;
        } else {
            r = m - 1;
        }
    }
    cout << ans << ' ' << out << endl;
}