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 <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
using namespace std;

int sum(vector<vector<long>> &data, vector<int> &comp,long G, int n){
    if (n == 0){
        int ans = 0;
        for (int sub = 0; sub < comp.size(); sub++){
            long sum = 0;
            int ansi = 0;
            if (comp.at(sub) == 1){
                continue;
            }
            for (int i = 0; i < comp.size(); i++){
                if (comp.at(i) == 1) {
                    sum += (i + 1) * 100 * data.at(i).at(0) + data.at(i).at(1);
                    ansi += data.at(i).at(0);
                }
            }
            for (int i = 0; i < data.at(sub).at(0) - 1; i++){
                if (sum < G) {
                    sum += (sub + 1) * 100;
                    ansi++;
                }
                else {
                    break;
                }
            }
            if (sum < G) {
                ansi = 0;
            }
            if (ans ==  0 || ansi < ans) {
                ans = ansi;
            }
        }
        return ans;
    }
    else {
        comp.at(comp.size() - n) = 0;
        int ans0 = sum(data, comp, G, n - 1);
        comp.at(comp.size() - n) = 1;
        int ans1 = sum(data, comp, G, n - 1);
        if (ans0 == 0){
            return ans1;
        }
        else if (ans1 == 0){
            return ans0;
        }
        return min(ans0, ans1);
    }
}

int main() {
    long D, G;
    cin >> D >> G;
    vector<vector<long>> data(D, vector<long>(2));
    for (int i = 0; i < D; i++) {
        for (int j = 0; j < 2; j++) {
            cin >> data.at(i).at(j);
        }
    }
    vector<int> comp(D, 0);
    int ans = sum(data, comp, G, D);
    cout << ans << endl;
}