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: Wrong Answer)

object Main {

  case class PUnit(i: Int, p: Int, c: Int) {
    def baseScores: Int = baseScore * p

    def baseScore: Int = 100 * (i + 1)

    def score: Int = baseScores + c

    def untilScore: Int = baseScore * (p - 1)

    def problems(score: Int): Int = score / baseScore
  }

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val D, G = sc.nextInt()
    val units = (0 until D).map { i =>
      val p, c = sc.nextInt()
      PUnit(i, p, c)
    }

    // コンプリートボーナスの比重が大きいので、当然コンプリートボーナスをとったほうがいい可能性
    // 問題得点は, 100点 ~ 1000点. 各最大100 x 100 = 10000, 1000 x 100 = 100000点程度
    // 問題数はたかだか10問.
    // * 基本的には全部解いて, ボーナスを獲得する. これを繰り返す.
    // どの問題から解いていけばいいか..

    // コンプリートボーナスがない場合のことを考える.
    // 当然その場合, 点数が大きい順に解いていく必要がある.

    // ボーナスがある場合, ボーナスが常に1000点以上なら, どの問題でもその問題の点数以上であることが
    // わかるので, 点数が大きい順に解いていけばいい.

    // ただ, ボーナスの点数が100 <= というのが問題
    // 100 <= ci < 1000のボーナスでは, 問題の点数よりも低い可能性がある.
    // p_9:1,c_9:1000 p_10:1,c_10:100の場合, p_9を1問解くと1900点 p_10を1問解くと1100点で
    // 必ずしも先に大きい点数の問題をとけば言い訳ではない.

    // ナップサック問題の変形?
    // 全部の問題点 + コンプリートボーナス を単位
    // オーバーした分は, 問題文が入りきる分だけ.
    // 単位が大きい順に入れる

    // それとも, 全部詰め込んで, 問題数が多い順の優先度キューで, 多い順に出せたら出す.
    // 残った

    // 全て解くか, 全てとか無いかを選択する.
    val INF = Int.MaxValue / 100

    def solve(): Int = {
      def dfs(i: Int, score: Int, unsolved: List[Int]): Int =
        if (i < D && score + units(i).score < G)
          Math.min(
            dfs(i+1, score + units(i).score, unsolved) + units(i).p,
            dfs(i+1, score, i :: unsolved))
        else if (i < D) dfs(i+1, score, i :: unsolved)
        else if (unsolved.isEmpty && score >= G) 0
        else if (unsolved.nonEmpty) {
          unsolved.find(score + units(_).untilScore >= G).map { i =>
            units(i).problems(G - score)
          }. getOrElse (INF)
        } else INF
      dfs(0, 0, Nil)
    }

    println(solve())
  }

}