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

typedef long long ll;
typedef pair<int, int> P;

int MOD = 1e9 + 7;
ll INF = 1e18;

int main() {
    ll d, g; cin>>d>>g;
    vector<ll> p(d), c(d), sum(d);
    vector<double> ef(d);
    bool flag=false;
    for(ll i=0; i<d; i++) {
        cin>>p[i]>>c[i];
        sum[i] = p[i] * 100 * (i + 1) + c[i];
        ef[i] = (double)(sum[i] / p[i]);
        //cout << sum[i] << " " << ef[i] << endl;
        if(sum[i] >= g) flag = true;
    }

    if(flag){
        ll MIN=INF;
        for(ll i=0; i<d; i++){
            ll g2=g, ans=0;
            
            for(ll j=1; j<=p[i]; j++){
                ans++;
                g2 -= 100 * (i + 1);

                if(j == p[i]) g2 -= c[i];

                if(g2 <= 0) break;
            }

            MIN = min(MIN, ans);
        }

        cout << MIN << endl;
    }

    else{
        ll ans2=0;
        while(1){
            double MAX=0;
            ll m;
            for(ll i=0; i<d; i++){
                if(MAX < ef[i]){
                    MAX = ef[i];
                    m=i;
                }
            }
            g -= sum[m];
            ans2 += p[m];
            sum[m]=0, p[m]=0, ef[m]=0;
           
            bool flag2=false;
            for(ll i=0; i<d; i++){
                if(sum[i] >= g) flag2 = true;
            }

            if(!flag2) continue;
            else{
                //cout << p[3]<<p[4]<<endl;
                //cout << g << endl;
                ll MIN=INF;
                for(ll i=0; i<d; i++){
                    ll g2=g, ans=INF;
                
                    for(ll j=1; j<=p[i]; j++){
                        if(j==1) ans=0;
                        ans++;
                        g2 -= 100 * (i + 1);

                        if(j == p[i]) g2 -= c[i];

                        if(g2 <= 0) break;
                    }

                    if(g2 > 0) ans = INF;

                    //cout << ans << endl;

                    MIN = min(MIN, ans);
                }
                cout << MIN + ans2 << endl;
                return 0;
            }
        }
    }
}