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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define newline cout<<"\n"
#define deb(x) cout<<#x<<" "<<x<<endl;

template<typename... T>
void read(T&... args){
    ((cin>>args), ...);
}

template<typename... T>
void write(T... args){
    ((cout<<args<<" "), ...);
}

void printarr(vector<ll> &v){
    for(ll x: v){
        cout<<x<<" ";
    }
    newline;
}

const ll MOD = 1e9+7;
const ll nax = 1e7+7;

int main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    
    ll n, g; read(n, g);

    vector<pair<ll, ll> > v(n);

    for(ll i=0; i<n; i++){
        cin>>v[i].first>>v[i].second;
    }

    ll ans = MOD;

    for(ll mask = 0; mask < 1<<n; mask++){
        ll tot = 0, cnt = 0, nt_max = -1;
        for(ll i=0; i<n; i++){
            if(mask>>i & 1){
                tot += 100*(i+1)*v[i].first + v[i].second;
                cnt += v[i].first;
            }
            else{
                nt_max = i;
            }
        }
        if(tot<g){
            ll req = (g-tot + (100*(nt_max+1))-1)/(100*(nt_max+1));
            if(req>=v[nt_max].first){
                continue;
            }
            else{
                cnt += req;
            }
        }
        ans = min(ans, cnt);
    }

    cout<<ans<<endl;
    
    return 0;
}