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<math.h>

int main(void){
  int d , g;
  scanf("%d %d",&d,&g);
  
  int p[d] , c[d];
  for(int i = 0;i < d;i ++){
    scanf("%d %d",&p[i],&c[i]);
  }
  int k = (int)pow(2,d);
  int kumi[k][d] , count = 0 , temp;
  
 for(int i = 0;i < k;i ++){
   count = 0;
   temp = i;
   while(count < d){
     if(temp%2!=0){
       kumi[i][count] = 1;
     }
     else if(temp%2==0){
       kumi[i][count] = 0;
     }
     count ++;
     temp /= 2;
   }
 }
  
  for(int i = 0;i < k;i ++){
    for(int j = 0;j < d;j ++){
      printf("%d ",kumi[i][j]);
    }
    putchar('\n');
  }
  
  
  int min = 1000000;
  int ten = 0 , ans = 0;
  
  for(int i = 0;i < k;i ++){
    for(int j = 0;j < d;j ++){
      if(kumi[i][j] == 1){
        ten += p[j]*100*(j+1)+c[j];
        ans += p[j];
      }
    }
    if(ten >= g){
      if(min > ans){
        min = ans;
      }
    }
    ans = 0;
    ten = 0;
  }
  
  printf("%d\n",min);
  
  return 0;
}