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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;

#define FOR(i,a,b) for (int i = a; i < b; ++i)
#define FORR(i,a,b) for (int i = b - 1; i >= a; --i)
#define REP(i,n) FOR(i,0,n)
#define REPR(i,n) FORR(i,0,n)

template<class T>
inline void read(T& x) {
  T s = 0, base = 1;
  char c;
  while (!isdigit(c=getchar())) if (c == '-') base = -base;
  while (isdigit(c)) {
    s = s * 10 + (c^48);
    c = getchar();
  }
  x = base * s;
}

int D, G;
int p[10], c[10];

int main() {
  read(D);
  read(G);
  G /= 100;

  REP(i,D) {
    read(p[i]);
    read(c[i]);
    c[i] /= 100;
  }

  int ans = numeric_limits<int>::max();
  for (int i = 0; i < (1 << D); ++i) {
    // i: 使いきったか否かを表すビット
    int sum = 0;
    int cnt = 0;

    for (int j = 0; j < D; ++D) {
      if ((i >> j) & 1) {
        sum += p[j] * (j + 1) + c[j];
        cnt += p[j];
      }
    }

    for (int j = D - 1; j >= 0; --j) {
      if ((i >> j) & 1) continue;
      if (sum >= G) break;
      int buf = min((G - sum) / (j + 1), p[j]);
      cnt += buf;
      sum += (j + 1) * buf;
    }

    if (G <= sum) ans = min(ans, cnt);
  }

  printf("%d\n", ans);
  return 0;
}