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;
#define endl '\n'

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);

    int n,k;

    cin >> n >> k;
vector<int> a(n),b(n);

    for(int i = 0;i < n;i++) cin >> a[i] >> b[i];

    bool t[10];
    int ans = INT_MAX;

    for(int i = 0;i < (1<<n);i++)
    {
        int tmp = i;
        int cnt = 0;
        for(int j = 0;j < n;j++) t[j] = false;
        while(tmp)
        {
            t[cnt] = tmp%2;
            tmp/=2;
            cnt++;
        }
        int each = 0,cn = 0;
        for(int j = 0;j < n;j++)
        {
            if(t[j])
            {
                cn+=a[j];
                each+=(j+1)*a[j]*100+b[j];
            }
        }
        if(cn>ans) continue;
        if(each>=k){ ans = min(ans,cn); continue; }
        for(int j = n-1;j >= 0;j--) if(!t[j])
        {
            for(int l = 0;l < a[j];l++)
            {
                each+=(j+1)*100;
                cn++;
                if(each>=k){ ans = min(ans,cn); break; }
            }
            break;
        }
    }

    cout << ans;
    cout << endl;
}