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>
double c[11][4];
void merge(int i,int j,double s[][4]){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d,k;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        for(k=0;k<4;k++)
          c[d][k]=s[b][k];
        b++;
      }
      else if(j<b){
        for(k=0;k<4;k++)
          c[d][k]=s[a][k];
        a++;
      }
      else if(s[a][3]>s[b][3] ||(s[a][3]==s[b][3] && s[a][0]<s[b][0])){
        for(k=0;k<4;k++)
          c[d][k]=s[a][k];
        a++;
      }
      else{
        for(k=0;k<4;k++)
          c[d][k]=s[b][k];
        b++;
      }
    }
    for(d=i;d<=j;d++)
      for(k=0;k<4;k++)
        s[d][k]=c[d][k];
  }
}

int main(){
  int d,g,count=0,i;
  scanf("%d%d",&d,&g);
  double p[d+1][4];
  for(i=1;i<=d;i++){
    p[i][0]=100*i;
    scanf("%lf%lf",&p[i][1],&p[i][2]);
    p[i][3]=i*100;
    p[i][3]+=p[i][2]/p[i][1];
  }
  merge(1,d,p);
  int num=1;
  while(g>p[num][0]*p[num][1]+p[num][2]){
    g-=p[num][0]*p[num][1]+p[num][2];
    count+=p[num][1];
    num++;
  }
  int min=100;
  for(i=num;i<=d;i++){
    int a=-1;
    if(g<p[i][0]*p[i][1]){
      a=g;
      if(a%(int)p[i][0]==0)
        a/=p[i][0];
      else
        a/=p[i][0]+1;
    }
    else if(g<p[i][0]*p[i][1]+p[i][2])
      a=p[i][1];
    if(min>a && a!=-1)
      min=a;
  }
  printf("%d\n",min+count);
  return 0;
}