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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"sort"
	"strconv"
)

func out(x ...interface{}) {
	fmt.Println(x...)
}

var sc = bufio.NewScanner(os.Stdin)

func getInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func getString() string {
	sc.Scan()
	return sc.Text()
}

// min, max, asub, absなど基本関数
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func asub(a, b int) int {
	if a > b {
		return a - b
	}
	return b - a
}

func abs(a int) int {
	if a >= 0 {
		return a
	}
	return -a
}

func lowerBound(a []int, x int) int {
	idx := sort.Search(len(a), func(i int) bool {
		return a[i] >= x
	})
	return idx
}

func upperBound(a []int, x int) int {
	idx := sort.Search(len(a), func(i int) bool {
		return a[i] > x
	})
	return idx
}

func main() {
	sc.Split(bufio.ScanWords)
	D, G := getInt(), getInt()
	p, c := make([]int, D), make([]int, D)
	s := make([]int, D)
	for i := 0; i < D; i++ {
		p[i], c[i] = getInt(), getInt()
		s[i] = 100*(i+1)*p[i] + c[i]
	}

	N := 1 << uint(D)
	ans := math.MaxInt64
	for i := 0; i < N; i++ {
		sum := 0
		cnt := 0
		for j := 0; j < D; j++ {
			if (i>>uint(j))%2 == 1 {
				sum += s[j]
				cnt += p[j]
			}
		}
		if sum >= G {
			for j := 0; j < D; j++ {
				cnt2 := 0
				sum2 := sum
				if (i>>uint(j))%2 == 1 {
					sum2 -= c[j]
					if sum2 <= G {
						break
					}
					for k := 0; k < p[j]; k++ {
						sum2 -= 100 * (j + 1)
						if sum2 <= G {
							sum2 += 100 * (j + 1)
							break
						}
						cnt2++
					}
				}
				// out(strconv.FormatInt(int64(i), 2), "---")
				// out(cnt, sum, cnt2, sum2)
				ans = min(ans, cnt-cnt2)
			}
			// out(sum)
		}
	}
	out(ans)
}