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 REP(i,n)	for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPS(i,s,n)	for(int i=(s), i##_len=(n); i<i##_len; ++i)

#define ALL(obj)	(obj).begin(),(obj).end()

typedef unsigned int 			uint;
typedef unsigned long long int	ull;
typedef long long int 			ll;

typedef pair<int,int>			P;

#define LINF		((ll)1 << 63 -1)
#define INF 		2000000007
#define MINF		-2000000007
#define MAX 		100005

const int MOD = 1e9+7;
// テンプレ終了
//====================================================================//

int D,G;
vector<int> p,c;

int func(int n){
		int suml=0;
		int cnt=0;
		REP(i,D){
			if(n&(1<<i)){
				cnt += p[i];
				suml += ((i+1)*100*p[i] + c[i]);
			}
		}
		if(suml < G){
			for(int i=D-1;i>=0;--i){
				if(!(n&(1<<i))){
					int diff = G-suml;
					int plus = (int)ceil((double)diff / ((i+1)*100));
					if(plus < p[i]){
						suml += plus*((i+1)*100);
						cnt += plus;
						break;
					}
					//cout << "\t" << diff << " " << plus << endl;
				}
			}
		}
		if(suml < G) cnt = INF;
	//cout << n << " " << cnt << " " << suml << endl;
	if(n&(1<<(D-1)))return cnt;
	if(n==0)return min(cnt,func(1));
	return min(min(cnt,func(n*2)),func(n*2+1));
}

// abc075
int main(){
	cin>>D>>G;
	p.resize(D);c.resize(D);
	REP(i,D){
		cin>>p[i]>>c[i];
	}
	
	int ans = INF;

	cout << func(0) << endl;

	return 0;
}