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;
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll inf=1e9+7;
const ll mod=998244353;
int dp[11][600000];
int main(){
    ll d;cin>>d;
    ll G;cin>>G;
    vector<ll>pt(1<<d);
    vector<pll>pc(d);
    vector<ll>cnt(1<<d);
    rep(i,d){
        ll p,c;
        cin>>p>>c;
        pc[i]=mp(p,c);
    }
    rep(i,1<<d){
        rep(j,d){
            if((1<<j)&i){
                pt[i]+=(j+1)*100*pc[j].first+pc[j].second;
                cnt[i]+=pc[j].first;
            }
        }
    }
    ll mi=inf;
    rep(i,1<<d){
        if(pt[i]>=G){
            mi=min(cnt[i],mi);
        }else{
            ll k=cnt[i];
            for(int j=d-1;j>=0;j--){
                if(!((1<<j)&i)){
                    if((pc[j].first-1)*(j+1)*100+pt[i]>=G){
                        k+=(G-pt[i])/((j+1)*100);
                        mi=min(k,mi);
                    }
                    break;
                }
            }
        }
    }
    cout<<mi<<endl;
}