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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
#define int long long

signed main() {
  int d, g;
  cin >> d >> g;
  int p[d], c[d];
  for (int i = 0; i < d; i++) {
    cin >> p[i] >> c[i];
  }
  int complete[d];
  for (int i = 0; i < d; i++) {
    int sum = (i + 1) * 100 * p[i] + c[i];
    complete[i] = sum;
  }
  int problem = 0;
  int point = 0;
  bool used[d];
  fill(used, used + d, false);
  for (int i = d - 1; i >= 0; i--) {
    if (point >= g) break;
    bool flag = false;
    for (int j = 0; j < d; j++) {
      if (i == j) continue;
      if (used[j]) continue;
      if (complete[j] / p[j] > 100 * (i + 1)) {
        flag = true;
      }
    }
    if (!flag) {
      int res = (g - point) / ((i + 1) * 100);
      if ((g - point) % ((i + 1) * 100)) res++;
      if (res < p[i] && res * (i + 1) * 100 + point >= g) {
        problem += res;
        point += res * (i + 1) * 100;
      } else {
        problem += p[i];
        point += complete[i];
      }
      used[i] = true;
    } else {
      int index = 0;
      double maxi = 0;
      for (int j = 0; j < d; j++) {
        if (i == j) continue;
        if (used[j]) continue;
        if (complete[j] / (double) p[j] > maxi) {
          maxi = complete[j] / (double) p[j];
          index = j;
        }
      }
      int res = (g - point) / ((i + 1) * 100);
      if ((g - point) % ((i + 1) * 100)) res++;
      if (res < p[index]) {
        point += (i + 1) * 100 * res;
        problem += res;
        used[i] = true;
      } else if (complete[index] + point >= g) {
        point += complete[index];
        problem += p[index];
        used[index] = true;
        i++;
      } else {
        int index = 0;
        int maxi = complete[0];
        for (int i = 1; i < d; i++) {
          if (maxi <= complete[i]) {
            index = i;
            maxi = complete[i];
          }
        }
        problem += p[index];
        point += complete[index];
        used[index] = true;
        if (i != index) i++;
      }
    }
  }
  cout << problem << endl;
  return 0;
}