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"
)

func main() {

	scanner := bufio.NewScanner(os.Stdin)
	var D, G int
	scanner.Scan()
	fmt.Sscanf(scanner.Text(), "%d %d", &D, &G)

	ps := make([]int, D)
	cs := make([]int, D)
	for i := 0; i < D; i++ {
		scanner.Scan()
		fmt.Sscanf(scanner.Text(), "%d %d", &ps[i], &cs[i])
	}

	ans := math.MaxInt64
	for index := uint(0); index < 1<<uint(D); index++ {
		targets := make(map[int]struct{})
		for shift := uint(0); shift < uint(D); shift++ {
			if ((index >> shift) & 1) == 1 {
				targets[int(shift)] = struct{}{}
			}
		}

		count, cost := 0, 0
		// get bonus
		for i, _ := range targets {
			cost = cost + ps[i]*100*(i+1) + cs[i]
			count = count + ps[i]
		}
		// without bonus
	B:
		for i := D - 1; i >= 0; i-- {
			if _, e := targets[i]; e {
				continue
			}
			for j := 0; j < ps[i]; j++ {
				cost = cost + 100*(i+1)
				count++
				if cost >= G {
					break B
				}
			}
		}
		if cost >= G {
			ans = min(ans, count)
		}
	}
	fmt.Println(ans)

}

func min(ns ...int) int {
	ret := math.MaxInt64
	for _, n := range ns {
		if n < ret {
			ret = n
		}
	}
	return ret
}