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: Accepted)

#include <bits/stdc++.h>
#define INF 9223372036854775807
using namespace std;
long long ans;
long long D,G,p,c;
vector<long long>vec_p;
vector<long long>vec_c;
long long solve(vector<long long>tmp_p, vector<long long>tmp_c) {
  long retans;
  long num,s,si,restmax, need;
  retans = INF;
  for(int mask=0; mask < (1 << D); mask++) {
    num = 0;
    s = 0;
    si = 0;
    restmax = 0;
    need = 0;
    for(int i = 0; i < D; i++) {
      if((mask >> i) & 1) {
        s += (i + 1) * 100 * tmp_p[i] + tmp_c[i];
        num += tmp_p[i];
      } else {
        restmax = i;
      }
    }
    if(s < G) {
      si = (restmax + 1) * 100;
      need = (G - s + si - 1) / si;
      if(need >= tmp_p[restmax]) {
        continue;
      }
      num += need;
    }
    retans = min(retans, num);
  }
  return retans;
}

int main(int arg, char* argv[]) {
  scanf("%d %d ", &D, &G);
  for(int i = 0; i < D; i++) {
    p = c = 0;
    scanf("%d %d", &p, &c);
    vec_p.push_back(p);
    vec_c.push_back(c);
  }
  ans = solve(vec_p, vec_c);
  cout << ans << endl;
  return 0;
}