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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <stdlib.h>
#include <fstream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <iterator>
#include <iomanip>
using namespace std;

int main(){
    long long d, g;
    cin >> d >> g;
    vector<pair<long long, long long> > a(d + 1);
    for(int i = 1; i <= d; ++i) cin >> a[i].first >> a[i].second;
    vector<vector<long long> > dp(d + 1, vector<long long>(1001));
    for(int i = 1; i <= d; ++i)
        for(int j = 1; j <= 1000; ++j){
            if(j < a[i].first) dp[i][j] = max(dp[i - 1][j], max(dp[i - 1][j - 1], dp[i][j - 1]) + 100 * i);
            else if(j == a[i].first) dp[i][j] = max(dp[i - 1][j], max(dp[i - 1][j - 1], dp[i][j - 1]) + 100 * i + a[i].second);
            else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); 
        }
    for(int i = 1; i <= 1000; ++i){
        if(dp[d][i] >= g){
            cout << i;
            return 0;
        }
    }
}