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 int long long
#define itn int
#define INF (1<<29)
#define rep(i,n) for(int i=0;i<(n);i++)
int gcd(int a,int b){
	if(b==0)return a;
	return gcd(b,a%b);
}
int lcm(int a,int b){
	return a/gcd(a,b)*b;
}
bool prime(int p){
	for(int i=2;i<=sqrt(p);i++){
		if(!p%i)return false;
	}
	return true;
}
//ライブラリ終了
int d,g,p[10],c[10],ans=INF,pr,cnt;
signed main(){
	cin>>d>>g;
	rep(i,d)cin>>p[i]>>c[i];
	for(int i=0;i<(1<<d);i++){
		pr=0,cnt=0;
		rep(j,d){
			if((i>>j)&1){
				cnt+=p[j]*(j+1)*100+c[j];
				pr+=p[j];
			}
		}
		for(int j=d-1;j>=0;j--){
			if(!(i>>j)&1){
				if(g>cnt){
					rep(k,p[j]){
						cnt+=(j+1)*100;
						pr++;
						if(g<=cnt)goto lop;
					}
					cnt+=c[j];
					if(g<=cnt)goto lop;
				}
				else goto lop;
			}
		}
		lop:;
		ans=min(ans,pr);
	}
	cout<<ans<<endl;
	return 0;
}