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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    static int modP = 1000000007;
    static long inf  = 1 << 61;
    static int[] x4 = {-1, 0, 0, 1};
    static int[] y4 = {0, -1, 1, 0};

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int d = in.nextInt();
        int g = in.nextInt();
        int[] P = new int[11];
        int[] C = new int[11];
        int[] S = new int[11];
        for(int i = 1; i <= d; i++) {
            P[i] = in.nextInt();
            C[i] = in.nextInt();
            S[i] = (P[i]*i*100) + C[i];
        }

        int ans = Integer.MAX_VALUE;
        for(int i = 0; i < (1 << d); i++) {
            int pointSum = 0;
            int problemCnt = 0;
            for(int j = 0; j < d; j++) {
                if(((i>>j)&1) == 1) {
                    pointSum += S[j+1];
                    problemCnt += P[j+1];
                }
            }

            if(pointSum >= g) {
                ans = Math.min(ans, problemCnt);
            }else{
                int pointsToGo = g - pointSum;

                for(int j = 0; j < d; j++) {
                    if(((i>>j)&1) == 0){
                        if(pointsToGo > (P[j+1]-1) * (j+1)*100) {
                            continue;
                        }else{
                            problemCnt += pointsToGo / ((j+1)*100);
                            ans = Math.min(ans, problemCnt);
                        }
                    }
                }
            }
        }

        System.out.println(ans);
    }
}