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

int score(vector<long long> &p, vector<long long> &c, long long &D, long long &G) {
  long long a = 0, b = 101;
  for(long long i = 0; i < D; i++) {
    if(G <= 100 * (i + 1) * p.at(i) + c.at(i)) {
      if(G / (100 * (i + 1)) + 1 <= p.at(i)) {
        a = G / (100 * (i + 1));
        if(G % (100 * (i + 1)) != 0) {
          a += 1;
        }
        
      }
      
      else{
      a = p.at(i);
      }
      
      if(a < b) {
        b = a;
      }
      
    }
    
  }
  
  if(b != 101) {
    return b;
  }
  
  long long x = 0, y = 0, t = 0, s = 0;
  for(long long i = 0; i < D; i++) {
    if(p.at(i) == 0) {
      x = 0;
    }
    
    else {
      x = ((100 * (i + 1) * p.at(i) + c.at(i)) / p.at(i) + (100 * (i + 1) * p.at(i) + c.at(i)) % p.at(i));
    }
    
    //cout << x << ":x" << i << endl;
    if(x > y) {
      y = x;
      t = i;
      s = p.at(i);
      //cout << y << ":y" << t << ":t" << s << ":s" << endl;
    }
    
  }
  G -= 100 * (t + 1) * p.at(t) + c.at(t);
  p.at(t) = 0;
  c.at(t) = 0;
  
  //cout << a << ":a " << b << ":b " << t << ":t " << G << ":G " << p.at(t) << ":p(t)" << endl;
  return (s + score(p, c, D, G));
}
  
  

int main() {
  long long D, G;
  cin >> D >> G;
  //cout << D << G << endl;
  vector<long long> p(D), c(D);
  for(long long i = 0; i < D; i++) {
    cin >> p.at(i) >> c.at(i);
    //cout << i << p.at(i) << c.at(i) << endl;
  }
  //一問を中途半端or完璧に解くケース
  long long s = 0, t = 101;
  //cout << s << t;
  for(long long i = 0; i < D; i++) {
    if(G <= (100 * (i + 1) * p.at(i) + c.at(i))) {
      if(G / (100 * (i + 1)) + 1 <= p.at(i)) {
        s = G / (100 * (i + 1));
        if(G % (100 * (i + 1)) != 0) {
          s += 1;
        }
        
      }
      
      else{
        s = p.at(i);
      }
      
      if(s <= t) {
        t = s;
      }
      
    }
    
  }
  //cout << s << t << endl;
  //tはこのケースにて最小の数となる。
  if(t != 101) {
    cout << t << endl;
  }
  
  if(t == 101) {
    cout << score(p, c, D, G) << endl;
  }
  
}