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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (d, g) = readLine()!!.split(" ").map { it.toInt() }

    // 通常スコアとコンプリートスコアがある
    // 問題を解くことで目標G 以上のスコアを獲得するのに解かなくてはいけない最小問題数は？

    // どのコンプボーナスを取るのか -> d < 10 -> 2^10 パターン
    // 必要なコンプボーナスを取って，後はスコアが高い順に上から埋めていけば OK
    // 2^10 パターンで G 以上となったときの問題数が一番少ないケースを求める

    val p = LongArray(d)
    val c = LongArray(d)
    for (i in 0 until d) {
        val (a, b) = readLine()!!.split(" ").map { it.toLong() }
        p[i] = a
        c[i] = b
    }

    var ans = Long.MAX_VALUE

    // パターンをどうやって表現するか？？
    for (i in 0..(Math.pow(2.0, d.toDouble()).toInt()-1) ) {

        // 0 埋めした２進数表現を取得
        val binaryRep = Integer.toBinaryString(i).toInt()
        val bs = "%0${d}d".format(binaryRep)

        val pTmp = p.clone()
        val cTmp = c.clone()

        var score = 0L
        var cnt = 0L
        // 使い切る問題を使い切る
        for ((j, ch) in bs.withIndex()) {
            if (ch == '1') {
                score += (j + 1) * 100 * pTmp[j] + cTmp[j]
                cnt += pTmp[j]
                pTmp[j] = 0
            }
        }

        // 残った問題を足していく
        // すべて使いきらない問題に対して大きい方から使う
        for (j in bs.lastIndex downTo 0) {
            if (score >= g) break

            if (bs[j] == '0') {
                // 問題を１問残すところまで使い切る
                while (score < g && pTmp[j] != 0L) {
                    score += (j + 1) * 100
                    cnt += 1
                    pTmp[j] -= 1L
                }
            }
        }

        if (score >= g) {
            ans = if (cnt < ans) cnt else ans
        }
    }

    println(ans)
}