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>

typedef struct{
  int p;
  int c;
  int pt;
}problem;

int comp(const void * a, const void *b)
{
  if(((problem *)a)->p < ((problem *)b)->p)
    return -1;
  else if(((problem *)a)->p == ((problem *)b)->p)
    return 0;
  else
    return 1;
}

int main(void)
{
  int d,g;
  problem probs[10];
  int i, j, k, buf, c1 = 0, c2 = 0,c3 = 0;

  scanf("%d %d",&d, &g);

  for(i = 0; i < d ; i++){
    scanf("%d %d",&probs[i].p, &probs[i].c);
    probs[i].pt = i;
  }

  g /= 100;
  for(i = 0; i < d; i++)
    probs[i].c /=  100;

  buf = g;
  for(i = d-1; i >= 0; i --){
    for(j = 0; j < probs[i].p; j ++){
      if (buf <= 0)
        goto out1;
     c1 ++; buf -= ( i + 1)  ;
    }
    buf -= probs[i].c;
  }
out1 :

  buf = g;
  for(i = 0; i < d; i ++){
    for (j = 0; j < probs[i].p; j ++){
      if(buf <= 0)
        goto out2;
      c2 ++; buf -= ( i + 1);
    }
    buf -= probs[i].c;
  }
out2 :
  
  qsort(probs,(size_t)d,sizeof(probs[0]),comp);
  buf = g;
  for(i = 0;i < d; i ++){
    for(j = 0; j < probs[i].p; j ++){
      if(buf <= 0)
        goto out3;
      c3 ++; buf -= (probs[i].pt +1);
    }
    buf -= probs[i].c;
  }

out3:
  
  if(c1 < c2 && c1 < c3)
    printf("%d\n",c1);
  else if(c2 < c3)
    printf("%d\n",c2);
  else
    printf("%d\n",c3);

  return(0);
}