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) = readInts()
  val PC = (1..D).map { readInts() }

  var min = Int.MAX_VALUE
  val b = Array(D) { false }

  fun f(): Int {
    var n = 0
    var g = G
    for (i in 0 until D) {
      if (b[i]) {
        val (p, c) = PC[i]
        val base = (i + 1) * 100
        n += p
        g -= c
        g -= p * base
      }
    }
    if (g <= 0) {
      return n
    }
    for (i in D - 1 downTo 0) {
      if (!b[i]) {
        val (p, c) = PC[i]
        val base = (i + 1) * 100
        val v = Math.min(g / base, p)
        n += v
        g -= v * base
        if (g <= 0) {
          return n
        }
      }
    }
    return n
  }

  fun dfs(i: Int) {
    if (i == D) {
      min = Math.min(f(), min)
    } else {
      b[i] = true
      dfs(i + 1)
      b[i] = false
      dfs(i + 1)
    }
  }
  dfs(0)

  println(min)
}

fun myBinarySearch(begin: Int, end: Int, isOk: (key: Int) -> Boolean): Int {
  var ng = begin
  var ok = end

  while (Math.abs(ok - ng) > 1) {
    val mid = (ok + ng) / 2
    if (isOk(mid)) {
      ok = mid
    } else {
      ng = mid
    }
  }

  return ok
}

fun <T> List<T>.myBinarySearch(isOk: (T) -> Boolean): Int {
  return myBinarySearch(-1, size) { index -> isOk(get(index)) }
}

fun <T> Array<T>.myBinarySearch(isOk: (T) -> Boolean): Int {
  return myBinarySearch(-1, size) { index -> isOk(get(index)) }
}

fun readString() = readLine()!!
fun readStrings() = readString().split(" ")
fun readInt() = readString().toInt()
fun readLong() = readString().toLong()
fun readInts() = readStrings().map { it.toInt() }
fun readLongs() = readStrings().map { it.toLong() }