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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int Inf = 1000000000;
int minmin = Inf;
int D, G;
vector<int> p;
vector<int> c;
map<pair<int, vector<int>>, int> dfsmap;

int dfs(int sum, vector<int> sp, int count) {
  if (dfsmap.count(make_pair(sum, sp)))
    return dfsmap[make_pair(sum, sp)];
  if (count > minmin)
    return Inf;
  if (sum >= G) {
    if (count < minmin)
      minmin = count;
    return 0;
  }
  else {
    int min_c = Inf;
    int s;
    vector<int> nsp;
    for (int i = D - 1; i >= 0 ; i--) {
      nsp = sp;
      if (nsp.at(i) == p.at(i))
        continue;
      nsp.at(i)++;
      if (nsp.at(i) == p.at(i))
        s = dfs(sum + (i + 1) * 100 + c.at(i), nsp, count + 1);
      else
        s = dfs(sum + (i + 1) * 100, nsp, count + 1);
      if (s < min_c)
        min_c = s;
    }
    return dfsmap[make_pair(sum, sp)] = 1 + min_c;
  }
}

int main() {
  cin >> D >> G;
  p = vector<int>(D);
  c = vector<int>(D);
  for (int i=0; i<D; i++)
    cin >> p.at(i) >> c.at(i);
  vector<int> sp(D, 0);
  cout << dfs(0, sp, 0) << endl;
}