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;
typedef long long ll;

int main(void) {
    ll D, G;
    cin >> D >> G;

    vector<pair<ll, ll>> p_c(D);
    for(int i = 0; i < D; i++) {
        cin >> p_c[i].first >> p_c[i].second;
    }

    vector<ll> ans_arr;

    for(int bit = 0; bit < (1 << D); bit++) {
        ll question_num = 0;
        ll total_point = 0;
        vector<ll> rest_index;

        for(ll i = 0; i < D; i++) {
            if(bit & (1 << i)) {
                pair<ll, ll> _p_c = p_c[i];
                total_point += 100 * (i + 1) * _p_c.first + _p_c.second;
                question_num += _p_c.first;
            } else {
                rest_index.push_back(i);
            }
        }

        if(total_point >= G) {
            ans_arr.push_back(question_num);
        } else {
            sort(rest_index.begin(), rest_index.end(), greater<ll>());
            for(ll i : rest_index) {
                int current_point = 100 * (i + 1);

                // 切り上げ
                int need_num = (G - total_point + current_point - 1) / current_point;

                pair<ll, ll> _p_c = p_c[i];
                if(need_num <= _p_c.first) {
                    ans_arr.push_back(question_num + need_num);
                    break;
                }

                ll solved_num = _p_c.first - 1;
                question_num += solved_num;
                total_point += current_point * solved_num;
            }
        }
    }

    sort(ans_arr.begin(), ans_arr.end());

    cout << ans_arr[0] << endl;
}