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)

import java.util.*
import java.math.*
import java.lang.Math.*

private val readString: ()->String = {readLine()!!}
private val readInt: ()->Int = { readLine()!!.toInt() }
private val readLong: ()->Long = { readLine()!!.toLong() }
private val readListInt: ()->List<Int> = { readLine()!!.split(' ').map(String::toInt) }
private val readListLong: ()->List<Long> = { readLine()!!.split(' ').map(String::toLong) }
private val errPrintln: (String)->Unit = { msg -> System.err.println(msg) }
private val MOD = 1e9.toLong()+7


fun greedyBitFS(item: List<Pair<Int, Int>>, bonus: List<Int>, LowLimit: Int): Int {
    val N = item.size
    var res = Int.MAX_VALUE;
    for (p in 0 until (0b01 shl N)) {
        var ptn = p
        var scr = 0
        var cnt = 0
        var lv = -1
        for(i in 0 until N) {
            val vi = item[i].first
            val ci = item[i].second
            if ((ptn and 0b01) == 1) {
                scr += vi * ci + bonus[i]
                cnt += ci;
                if (scr >= LowLimit) {
                    res = min(res, cnt)
                }
            } else {
                lv = max(lv, i)
            }
            ptn = ptn shr 1
        }
        if (scr < LowLimit) {
            if (lv >= 0) {
                val v = item[lv].first
                val c = item[lv].second
                for(j in c-1 downTo  1) {
                    if (scr + v * j >= LowLimit) {
                        res = min(res, cnt + j)
                    }
                }
            }
        }
    }
    return res
}


fun main(args: Array<String>) {
    val (D,G) = readListInt()
    val item = Array<Pair<Int, Int>>(D){ Pair(0,0) }
    val C = Array<Int>(D){0}

    for(i in 0 until D) {
        val pc = readListInt()
        item[i] = Pair(100*(i+1), pc[0])
        C[i] = pc[1]
    }

    val ans = greedyBitFS(item.toList(), C.toList(), G)

    println(ans)
}