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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

void get_intint(char *line, int size, int *a1, int *a2) {
  if(!fgets(line, size, stdin)) return;
  sscanf(line, "%d %d", a1, a2);
}

#define BUF_SIZE 20


struct problem {
  int point;
  int bonus;
  int cnt;
  int sum;
};

int solve_min(struct problem* probs, int grades, int goal) {
  int mincnt = INT_MAX;
  int i, j;
  for(i = 0; i < (1<<grades); i++) {
    int sum = 0;
    int tmpcnt = 0;
    // get bonus
    for(j = 0; j < grades; j++) {
      int flag = !!(i & (1<<j));
      tmpcnt += probs[j].cnt*flag;
      sum += probs[j].sum*flag;
    }
    if(tmpcnt > mincnt) continue; // skip
    if(sum >= goal) {
      mincnt = tmpcnt;
      break;
    }
    // append reminder
    for(j = grades-1; j >= 0; j--) {
      int flag = !!(i & (1<<j));
      if(flag) continue; // it would get bonus twice
      int rem = (goal-sum+probs[j].point-1) / probs[j].point;
      if(rem >= probs[j].cnt) continue;
      tmpcnt += rem;
      if(mincnt > tmpcnt) mincnt = tmpcnt;
      break;
    }
  }
  return mincnt;
}

int main(void) {
  int grade, total_score;
  char line[BUF_SIZE];
  get_intint(line, BUF_SIZE, &grade, &total_score);

  struct problem probs[10];

  int i, j;
  int max_cnt = 0;
  for(i = 0; i < grade; i++) {
    int cnt, bonus;
    get_intint(line, BUF_SIZE, &cnt, &bonus);
    probs[i].cnt = cnt;
    probs[i].bonus = bonus;
    probs[i].point = (i+1)*100;
    probs[i].sum = probs[i].point * probs[i].cnt + bonus;
    max_cnt += probs[i].cnt;
  }

  int res = solve_min(probs, grade, total_score);
  printf("%d\n", res);
}