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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void){
	int D,G,ans=1e9,i;
	scanf("%d %d",&D,&G);
	int p[D+1],c[D+1];
	for(i=1;i<=D;i++)scanf("%d %d",&p[i],&c[i]);
	int list;
	//num << n でnumが2^n倍される
	//1 << D = 2^D
	for(list=0; list<(1<<D); list++){
		int score=0, num=0, rest_max;
		for(i=1; i<=D; i++){
			if(list>>(i-1) & 1){
			//listを左にiビットシフトし、1とのビット論理和
			//=> listの2進数表記のi桁目が１なら解く
				score += 100*i*p[i] + c[i];
				num += p[i];
			}
			else{
				rest_max = i;//解いていない
			}
		}
		if(score<G){
			int s1 = 100*rest_max;//中途半端に解く問題の点数
			int need = (G-score+s1-1)/s1;
			if(need > p[rest_max]) continue;//以下を実行せず次のループへ
			num += need;
		}
		ans = ans<num? ans: num;	
	}
	printf("%d",ans);	
	return 0;
}