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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

typedef struct{
    long point;
    int num;
    long comp_point;
}question;

int main(void){
    int D;
    long G;
    scanf("%d%ld",&D,&G);

    //Q[0]~Q[D-1]に格納する
    question* Q;
    NEW(Q,D);
    for(int i=0;i<D;i++){
        Q[i].point=(i+1)*100;
        scanf("%d%ld",&Q[i].num,&Q[i].comp_point);
    }

    int ans=1000000000;
    //tのiビット目が1→全部とく、0→全部はとかない
    for(int t=1;t<(1<<D);t++){
        int tmp=0;
        int pnt=0;
        //全部とくものを処理
        for(int i=0;i<D;i++){
            if(t&(1<<i)){
                tmp+=Q[i].num;
                pnt+=Q[i].point*Q[i].num+Q[i].comp_point;
            }
        }
        if(pnt>G) continue;

        //全部はとかないものを処理
        for(int i=D-1;i>=0;i--){
            if(!(t&(1<<i))){
                //その問題ではきっちりG点にできないもの
                if((G-pnt)%Q[i].point!=0) continue;
                //その問題を全部解いてもG点にできないもの
                else if((G-pnt)/Q[i].point>=Q[i].num) continue;
                //OKな場合は処理
                else{
                    tmp+=(G-pnt)/Q[i].point;
                    ans=MIN(ans,tmp);
                    break;
                }
            }
        }
    }
    printf("%d\n",ans);

    return 0;
}