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: Wrong Answer)

import java.util.*;
import java.io.*;

public class Main {
    static class Pair {
        long c;
        long bonus;
        Pair (long c, long bonus) {
            this.c = c;
            this.bonus= bonus;
        }
    }
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        Scanner sc= new Scanner(System.in);
        long d = sc.nextLong();
        long ans = Long.MAX_VALUE;
        long g = sc.nextLong();
        Pair[] arr = new Pair[(int)d];
        int flag = 0;
        for (int i=0; i<d; i++) {
            long a = sc.nextLong();
            long b = sc.nextLong();
            arr[i] = new Pair(a, b);
            long val = (100*(i+1) * a);
            if (val == g) {
                flag = 1;
                ans = Math.min(ans, a);
            }
            else if (val > g) {
                long total = 0l;
                long count = 0l;
                for (int j=0; j<a; j++) {
                    val -= (100*(i+1));
                    count++;
                    if (val == g) {
                        break;
                    } else if (val < g) {
                        count--;
                        break;
                    }
                }
                ans = Math.min(ans, a-count);
                flag = 1;
            } else if (val+b>=g) {
                ans = Math.min(ans, a);
                flag = 1;
            }
        }
        
        if (flag == 1) {
            //System.out.println(ans);
            long xx = ans;
            long sum = 0l;
            ans = 0l;
            for (long i=d-1; i>=0; i--) {
                Pair p = arr[(int)i];
                long count = p.c;
                long bonus = p.bonus;
                sum += (count*(100*(i+1)));
                //System.out.println("sum="+sum);
                if (sum == g) {
                    ans += count;
                    break;
                }
                else if (sum > g) {
                    long total = (100*(i+1));
                    long cc = 0;
                    for (int j=0; j<count; j++) {
                        //System.out.println("count="+count);
                        //System.out.println("tital="+total);
                        sum -= total;
                        cc++;
                        //System.out.println("g-"+g);
                        if (sum == g) {
                            //cc--;
                            break;
                        } else if (sum < g) {
                            cc--;
                            break;
                        }
                    }
                    //System.out.println("cc="+cc);
                    //if (cc==0) 
                    ans += count-cc;
                    //else ans+=count-cc-1;
                    break;
                } else if (sum+bonus >= g) {
                    ans += count;
                    break;
                } else {
                    //System.out.println("yo");
                    sum += bonus;
                    ans += count;
                    if (sum >= g) break;
                    
                }
            }
            
            System.out.println(Math.min(xx,ans));
        } else {
            long sum = 0l;
            ans = 0l;
            for (long i=d-1; i>=0; i--) {
                Pair p = arr[(int)i];
                long count = p.c;
                long bonus = p.bonus;
                sum += (count*(100*(i+1)));
                //System.out.println("sum="+sum);
                if (sum == g) {
                    ans += count;
                    break;
                }
                else if (sum > g) {
                    long total = 0;
                    long cc = 0;
                    for (int j=0; j<count; j++) {
                        total = (100*(i+1));
                        //System.out.println("count="+count);
                        //System.out.println("tital="+total);
                        sum -= total;
                        cc++;
                        //System.out.println("g-"+g);
                        if (sum <= g) {
                            //cc--;
                            break;
                        }
                    }
                    //System.out.println("cc="+cc);
                    //if (cc==0) 
                    ans += count-cc;
                    //else ans+=count-cc-1;
                    break;
                } else if (sum+bonus >= g) {
                    ans += count;
                    break;
                } else {
                    //System.out.println("yo");
                    sum += bonus;
                    ans += count;
                    if (sum >= g) break;
                    
                }
            }
            System.out.println(ans);
        }
            //if (ans == 0) ans = Long.MAX_VALUE;
            
        //
    }
}