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: Runtime Error)

fun main(args: Array<String>) {
    val (dt, g) = readLine()!!.split(' ').map(String::toLong)
    val d = dt.toInt()
    val pArray = Array(d + 1) { arrayOfNulls<Long>(101) }
    val scoreArray = Array<Long?>(1001) { null }
    scoreArray[0] = 0L
    for (i in 1..d) {
        val (p, c) = readLine()!!.split(' ').map(String::toLong)
        for (j in 1..p.toInt()) {
            pArray[i][j] = (i.toLong() * 100) * j
            if (j == p.toInt()) pArray[i][j] = pArray[i][j]!! + c
        }
    }

    for (i in 1..d) {
        var selectNum = 1
        val scoreArrayBefore = scoreArray.copyOf()
        while (selectNum != 101 || pArray[i][selectNum] != null) {
            for (k in 1..1000) {
                if (selectNum > k) continue
                val value = scoreArrayBefore[k - selectNum] ?: break
                if (scoreArray[k] == null ||
                        scoreArrayBefore[k] ?: 0 < (value + pArray[i][selectNum]!!)) {
                    scoreArray[k] = value + pArray[i][selectNum]!!
                }
                if (scoreArray[k] ?: 0 >= g) break
            }
            selectNum++
        }
    }
    for (i in 1..1000) {
        if (scoreArray[i] ?: 0 >= g) {
            println(i)
            break
        }
    }
}