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

#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
typedef long long unsigned int ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1}; // 移動方向

int d,g;
int p[11],c[11];

void input(){
    cin >> d >> g;
    for(int i=0;i<d;i++) cin >> p[i] >> c[i];
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    input();
    double ave[d];
    int rank[d];
    int count = 0;
    for(int i=0;i<d;i++){
        ave[i] = (100*p[i]*(i+1) + c[i]) / p[i];
    }
    sort(ave, ave+d, greater<double>()); // 平均値が大きい順に並べる
    for(int i=0;i<d;i++){
        for(int j=0;j<d;j++){
            if(ave[j] == ((100*p[i]*(i+1) + c[i]) / p[i])) rank[j] = i;
        }
    }
    for(int i=0;i<d;i++){
        for(int j=0;j<p[rank[i]];j++){
            g -= (rank[i]+1)*100;
            count++;
            if(g <= 0){
                cout << count << endl;
                return 0;
            }
        }
        if (g > 0){
            g -= c[rank[i]];
            if (g <= 0){
                cout << count << endl;
                return 0;
            }
        }
        else{
            cout << count << endl;
            return 0;
        }
    }
    cout << count << endl;
    return 0;
}