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: Accepted)

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


int main(void){
    //////////////////////////

    int D, G;
    cin >> D >> G;
    vector<int> p(D), c(D);
    for (int i = 0; i < D; i++){
        cin >> p.at(i) >> c.at(i);
    }
    int ans = 1000000000;
    for (int i = 0; i < pow(2, D); i++){
        int now_score = 0;
        int cnt = 0;
        vector<bool> check(D);
        for (int j = 0; j < D; j++){
            if (i & (1 << j)){
                now_score += 100 * (j+1) * p[j] + c[j];
                cnt += p[j];
                check[j] = true;
            }
        }
        if (now_score < G){
            for (int j = D-1; j >= 0; j--){
                if (!check[j]){
                    int z = (G - now_score + 100 * (j+1) - 1) / (100 * (j+1));
                    if (z > p[j]){
                        now_score += 100 * (j+1) * (p[j] - 1);
                        cnt += p[j] - 1;
                    }else{
                        now_score += 100 * (j+1) * z;
                        cnt += z;
                    }
                }
                if (now_score >= G){
                    break;
                }
            }
        }
        if (now_score >= G){
            ans = min(ans, cnt);
        }   
    }

    cout << ans << endl;

    //////////////////////////
    return 0;
}