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: Time Limit Exceeded)

import java.lang.Math.min

fun main(args: Array<String>) {

    val (d, g) = readLine()!!.split(' ').map(String::toInt)
    val scores = (1..d).map{
        val (p, c) = readLine()!!.split(' ').map(String::toInt)
        listOf(100 * it, p, c)
    }
    print(calcF1(g, scores.reversed(), 0))
}

fun calcF1(goal: Int, probs: List<List<Int>>, acc: Int): Int?{
    val len = probs.size
    fun inI(i : Int, goal: Int, bit: Int, acc: Int): Int?{
        if(goal < 0) return acc
        if(i !in probs.indices) {
            val ps = probs.mapIndexedNotNull{index, list ->  if(1 and (bit shr index) == 1) null else list}
            return calcF2(goal, ps, acc)
        }
        val iscore = probs[i].let { it[0] * it[1] + it[2]}
        return min(inI(i+1, goal - iscore, bit + (1 shl i), acc + probs[i][1]) ?: Int.MAX_VALUE, inI(i+1, goal, bit, acc) ?: Int.MAX_VALUE)
    }
    return inI(0, goal, 0, acc)
}

fun calcF2(goal: Int, probs: List<List<Int>>, acc: Int): Int?{
    if(goal <= 0) return acc
    if(probs.isEmpty()) return null

    val current = probs.first()
    val rest = probs.subList(1, probs.size)

    return (0..current[1]).mapNotNull {i ->
        val score = current[0] * i + if(i == current[1]) current[2] else 0
        calcF2(goal - score, rest, acc + i)
    }.min()
}