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: Runtime Error)

import java.util.*;
//https://atcoder.jp/contests/abc104/tasks/abc104_c

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        final int D = Integer.parseInt(sc.next());
        final int G = Integer.parseInt(sc.next())/100;
        int p[] = new int[D];
        int c[] = new int[D];
        int m[] = new int[D];
        int n[] = new int[D];
        for (int i=0; i<D; i++) {
            p[i] = Integer.parseInt(sc.next());
            c[i] = Integer.parseInt(sc.next())/100;
            m[i] = p[i]*(i+1);
            n[i] = m[i]+c[i];
        }
        sc.close();

        boolean f = false;
        int min = Integer.MAX_VALUE;
        for (int i=D-1; i>=0; i--) {
            if (m[i]>=G) {
                f = true;
                for (int j=0; j<p[i]; j++) {
                    if ((i+1)*(j+1)>=G && min>j) {
                        min = j;
                    }
                }
            }
        }
        for (int i=0; i<D; i++) {
            if (n[i]>=G) {
                f = true;
                if (min>p[i]+1)
                    min = p[i];
            }
        }
        if (f) {
            System.out.println(min);
            return;
        }
        int sum = n[D-1];
        int cnt = p[D-1];
        for (int i= D-2; i>=0; i--) {
            if (sum+n[i]==G) {
                System.out.println(cnt+p[i-1]);
                return;
            } else if (sum+n[i]>G) {
                for (int j=0; j<p[i]; j++) {
                    sum += i+1;
                    if (sum>=G) {
                        System.out.println(cnt+j+1);
                        return;
                    }
                }
                sum += c[i];
            } else {
                sum += n[i];
                cnt += p[i];
            }
        }
    }
}