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>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main(void) {
  int D, G;
  int p[10], c[10];
  scanf("%d %d", &D, &G);
  G /= 100;
  for (int i = 0; i < D; i++) {
    scanf("%d %d", &p[i], &c[i]);
    c[i] /= 100;
  }

  int all = 0, ans[10] = {}, min = 0x7FFFFFFF;
  while (all < (1 << D)) {
    int score = 0;
    for (int i = 0; i < D; i++) {
      ans[i] = ((all & (1 << i)) > 0) * p[i];
      score += ans[i]*(i+1) + (ans[i]>0)*c[i];
    }
    for (int i = D-1; G > score && i >= 0; i--) {
      if ((all & (1 << i)) > 0) continue;
      if (ceil((G - score) / (double)(i+1)) >= p[i]) {
        ans[i] = p[i];
        score += (i+1)*p[i] + c[i];
      } else {
        ans[i] = ceil(((double)G - score) / (i+1));
        score += (i+1)*ans[i];
      }
    }
    int n = 0;
    for (int i = 0; i < D; i++) n += ans[i];
    if (min >= n) {
      min = n;
    }
    all++;
  }
  printf("%d\n", min);

  return 0;
}