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<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<cstring>
#include<math.h>
#include<bitset>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
#include<assert.h>
#include<string>
#include<cstdlib>
#include<cctype>
#include<unordered_map>
#include<sstream>
// #include<bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int,int>;
constexpr ll INF = 1LL << 60;
const double INFD = 1e100;
const ll mod = 1000000007;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
};
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
};

// intやllをstringに変換
template<typename T> inline string toString(const T &a){
  ostringstream oss;
  oss << a;
  return oss.str();
};
// ----------------------------------------------------------------------------

int main(){
  ll D,G;
  cin >> D >> G;
  vector<ll> p(D),c(D);
  for(int i=0; i<D; i++){
    cin >> p[i] >> c[i];
  }
  ll ans = 1e5;
  for(int bit=0; bit<(1<<D); bit++){
    ll point = 0;
    ll cnt = 0;
    for(ll i=0; i<D; i++){
      if(bit & (1<<i)){
        cnt += p[i];
        point += c[i];
        point += (i+1)*100*p[i];
      }
    }
    if(point >= G){
      chmin(ans,cnt);
      continue;
    }
    for(ll i=D-1; i>=0; i--){
      if(!(bit & (1<<i))){
        if((i+1)*100*(p[i]-1)+point < G){
          point += (i+1)*100*(p[i]-1);
          cnt += p[i]-1;
        }else{
          ll num;
          num = (G-point+(i+1)*100-1)/((i+1)*100);
          cnt += num;
          point += (i+1)*100*num;
          break;
        }
      }
    }
    if(point >= G){
      chmin(ans, cnt);
    }
  }
  cout << ans << endl;
  return 0;
}