Problem:
Score : 300 points

Problem Statement

A programming competition site AtCode provides algorithmic problems.
Each problem is allocated a score based on its difficulty.
Currently, for each integer i between 1 and D (inclusive), there are p_i problems with a score of 100i points.
These p_1 + … + p_D problems are all of the problems available on AtCode.

A user of AtCode has a value called total score.
The total score of a user is the sum of the following two elements:

Base score: the sum of the scores of all problems solved by the user.

Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of c_i points, aside from the base score (1 ≤ i ≤ D).

Takahashi, who is the new user of AtCode, has not solved any problem.
His objective is to have a total score of G or more points.
At least how many problems does he need to solve for this objective?

Constraints

1 ≤ D ≤ 10

1 ≤ p_i ≤ 100

100 ≤ c_i ≤ 10^6

100 ≤ G

All values in input are integers.

c_i and G are all multiples of 100.

It is possible to have a total score of G or more points.

Input

Input is given from Standard Input in the following format:

D G
p_1 c_1
:
p_D c_D

Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in Scala (2.11.7) (judge verdict: Accepted)

object Main extends App {
  case class Problem(n: Int, ratio: Long, bonus: Long) {
    val completeScore = n * ratio + bonus
  }
  case class Result(score: Long, count: Int)

  val in = new java.util.Scanner(System.in)
  val D = in.nextInt
  val G = in.nextLong / 100

  val Ps = Array.tabulate(D){i => Problem(in.nextInt, i + 1, in.nextLong / 100)}

  def getAll(halfway: Int): Seq[Int] = {
    val comps: Seq[Result] = (0 until (1 << D)).flatMap {mask =>
      if (((1 << halfway) & mask) != 0) None
      else {
        Some((0 to D).foldLeft(Result(0L, 0)){case (r, i) =>
          if (((1 << i) & mask) == 0) r
          else Result(r.score + Ps(i).completeScore, r.count + Ps(i).n)
        })
      }
    }

    val p = Ps(halfway)
    val halfways: Seq[Result] = (0 to p.n).map {n =>
      if (n == p.n) Result(p.completeScore, n)
      else Result(n * p.ratio, n)
    }

    for {
      c <- comps
      h <- halfways
      if (c.score + h.score >= G)
    } yield c.count + h.count
  }

  println((0 until D).flatMap(getAll).min)
}