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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main ()
{
    int D, G;
    std::cin >> D >> G;
    
    int p[15] = {0};
    int c[15] = {0};
    
    for (int i = 1; i <= D; i++) 
    {
        std::cin >> p[i] >> c[i];
    }
    
    int max_i = D;
    //std::cout << "max number: " << max_i << std::endl;
    
    int point = 0;
    int count = 0;
    bool flag = false;
    
    //std::cout << "Strategy: 1" << std::endl;
    
    for (int i = max_i; i > 0; i--) 
    {
        //std::cout << "i: " << i << std::endl;
        
        for (int j = 1; j <= p[i]; j++) 
        {
            count++;
            point += 100*i;
            if (j == p[i])
            {
                point += c[i];
            }
            
            //std::cout << "count: " << count << std::endl;
            //std::cout << "point: " << point << std::endl;
            
            if (point >= G)
            {
                flag = true;
                //std::cout << "over G!" << std::endl;
            }
            
            if (flag)
            {
                break;
            }
        }
        
        if(flag)
            break;
    }

    //std::cout << "answer 1: " << count << std::endl;

    //std::cout << "Strategy: 2" << std::endl;

    int point2 = 0;
    int count2 = 0;
    bool flag2 = false;
    
    int mini = p[1];
    point2 = 100*p[1] + c[1];
    count2 = p[1];
    for (int i = 2; i <= D; i++) 
    {
        if (mini >= p[i])
        {
            point2 = 100*p[i] + c[i];
            count2 = p[i];
        }
        //std::cout << "count2: " << count2 << std::endl;
        //std::cout << "point2: " << point2 << std::endl;
    }
    
    //std::cout << "answer 2: " << count2 << std::endl;
    
    if (point2 >= G)
    {
        std::cout << min(count, count2) << std::endl;
    }
    else
    {
        std::cout << count << std::endl;
    }
    
    return 0;
}