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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
#include <iostream>
using namespace std;

int t[10];

int main()
{
  int d, g, p[10], c[10];
  cin >> d >> g;
  for(int i = 0; i < d; i++){
    cin >> p[i] >> c[i];
    t[i] = (i+1)*100*p[i] + c[i];
  }

  int ans = 1e+9;
  for(int i = 0; i < (1<<d); i++){
    int s = 0, cnt = 0, r = -1;
    for(int j = 0; j < d; j++){
      if(i >> j & 1){
        s += t[j];
        cnt += p[j];
      }else{
        r = j;
      }
    }

    if(s < g){
      int ss = (r+1)*100;
      int need = (g - s + ss - 1) / ss;
      if(need >= p[r]) continue;
      cnt += need;
    }

    ans = min(ans, cnt);
  }

  cout << ans << endl;

  return 0;
}
