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>
#define REP(i,n) for(int i=0;i<(n);i++)
#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REV(c) reverse((c).begin(),(c).end())

using namespace std;

// 最大公約数
int64_t gcd(int64_t a, int64_t b) {
  return b != 0 ? gcd(b, a % b) : a;
}

// 最小公倍数
int64_t lcm(int64_t a, int64_t b) {
  return a * b / gcd(a, b);
}

int ctoi(char c){
  return c - '0';
}

struct UnionFind {
  vector< int > data;
 
  UnionFind(int sz) {
    data.assign(sz, -1);
  }
 
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if(x == y) return (false);
    if(data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    return (true);
  }
 
  int find(int k) {
    if(data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
 
  int size(int k) {
    return (-data[find(k)]);
  }
};

int64_t rec(auto vec, int g, int i, int d, int s, int c){
  int t = s + i * (d+1) * 100;
  if (i == vec.at(d).first){
    t += vec.at(d).second;
  }
  if (d == vec.size()-1){
    if (t >= g)
      return c;
    else
      return INT_MAX;
  }else{
    int mi = INT_MAX;
    REP(j, vec.at(d+1).first+1){
      int tmp = rec(vec, g, j, d+1, t, c+j);
      mi = min(mi,tmp);
    }
    return mi;
  }
}

int main(){
  int64_t d,g;
  cin >> d >> g;
  vector<pair<int, int>> vec(d);
  REP(i,d){
    cin >> vec.at(i).first >> vec.at(i).second;
  }
  int mi = INT_MAX;
  REP(i,vec.at(0).first+1){
    int tmp = rec(vec, g, i, 0, 0, i);
    mi = min(mi,tmp);
  }
  cout << mi << endl;
  return 0;
}