You have decided to practice. Practice means solving a large number of problems in AtCoder.  
Practice takes place over several days. The practice for one day is carried out in the following steps.

-   Let $M$ be the number of problems to be solved on that day. Each problem has a value called difficulty, which is a pair of non-negative integers $(A, B)$.
-   First, rearrange the $M$ problems in any order. Then, solve the problems one by one in that order.
-   You have a value called **fatigue**. The fatigue at the beginning of the day is $0$, and it changes from $x$ to $Ax + B$ when solving a problem with difficulty $(A, B)$.
-   The fatigue at the end of solving all $M$ problems is called the energy consumed for that day.

There is a sequence of $N$ problems in AtCoder, called problem $1$, problem $2$, $\dots$, problem $N$ in order. The difficulty of problem $i$ is $(A_i, B_i)$.  
You have decided to solve all $N$ problems through practice.  
Practice is carried out in the following steps. Below, let $[L, R]$ denote the following sequence of problems: problem $L$, problem $L+1$, $...$, problem $R$.

-   Choose an integer $K$ freely between $1$ and $N$, inclusive. The practice will last $K$ days.
-   Divide the sequence of $N$ problems into $K$ non-empty continuous subsequences, and let $S_i$ be the $i$-th sequence.  
    Formally, choose a strictly increasing non-negative integer sequence $x_0, x_1, \dots, x_K$ such that $1 = x_0$ and $x_K = N + 1$, and let $S_i = [x_{i-1}, x_i - 1]$ for $1 \leq i \leq K$.
-   Then, for $i=1, 2, \dots, K$, solve all problems in $S_i$ on the $i$-th day of practice.

You have decided to practice so that the total energy consumed throughout the practice is at most $X$.  
Let $D$ be the minimum possible value of $K$, the number of days, for such a practice. (Here, it is guaranteed that $\sum_{i = 1}^N B_i \leq X$. Under this constraint, such $D$ always exists.)  
Also, let $M$ be the minimum possible total energy consumed throughout a practice such that $K=D$.

Find $D$ and $M$.

The input is given from Standard Input in the following format:

```
$N$ $X$
$A_1$ $B_1$
$A_2$ $B_2$
$\vdots$
$A_N$ $B_N$
```

Print $D$ and $M$ separated by a space.

-   $1 \leq N \leq 2 \times 10^5$
-   $1 \leq X \leq 10^8$
-   $1 \leq A_i \leq 10^5$
-   $1 \leq B_i$
-   $\sum_{i=1}^N B_i \leq X$
-   All input values are integers.

1 52

In this test case, you can solve all problems in one day while satisfying the condition for X.
By following the steps below, the total energy consumed during the practice will be 52, which is the minimum.

• Set K=1 and solve [1,3] on the first day.
• Carry out the practice on the first day in the following steps.
  • Arrange the problems in the order (problem 3, problem 2, problem 1).
  • Initially, the fatigue is 0.
  • Solve problem 3. The fatigue changes to 5×0+7=7.
  • Solve problem 2. The fatigue changes to 3×7+4=25.
  • Solve problem 1. The fatigue changes to 2×25+2=52.
  • The fatigue at the end of solving all problems is 52. Therefore, the energy consumed on this day is 52.
• The total energy consumed throughout the practice is also 52.

2 17

This test case is obtained by changing X from 100 to 30 in the first test case. Therefore, it is impossible to solve all problems in one day while satisfying the condition for X.
By following the steps below, you can achieve M=17 by practicing for two days.

• Set K=2, and solve [1,2] on the first day and [3,3] on the second day.
• Carry out the practice on the first day in the following steps.
  • Arrange the problems in the order (problem 1, problem 2).
  • Initially, the fatigue is 0.
  • Solve problem 1. The fatigue changes to 2×0+2=2.
  • Solve problem 2. The fatigue changes to 3×2+4=10.
  • The fatigue at the end of solving all problems is 10. Therefore, the energy consumed on the first day is 10.
• Carry out the practice on the second day in the following steps.
  • Arrange the problems in the order (problem 3).
  • Initially, the fatigue is 0.
  • Solve problem 3. The fatigue changes to 5×0+7=7.
  • The fatigue at the end of solving all problems is 7. Therefore, the energy consumed on the second day is 7.
• The total energy consumed throughout the practice is 10+7=17.

5 50000000

The optimal practice is to solve one problem per day.