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<algorithm>
#include<vector>
#include<set>
using namespace std;

int d,g,mini = 999999999;
vector<int> qn,bn;
vector<int> stb,st_nob;

void dfs4(int gr,vector<bool> sn) {
    if (gr == d) {
        int pt=0,cont=0;
        for (int i=0;i<d;i++) {
            if (sn[i]) {pt+=stb[i];cont+=qn[i];}
        }
        if (pt>g) { 
            if (mini>cont) mini = cont;
        }
        else {
            for (int j=d-1;j>=0;j--) {
                if (sn[j]==false) {
                    if (pt+st_nob[j]>=g) {
                        int sub = (g-pt)/((j+1)*100);
                        if ((g-pt)%((j+1)*100)!=0)++sub;
                        if (mini > cont+sub) mini = cont+sub;
                    }
                    break;
                }
            }
        }
    }
    else {
        vector<bool> tmp1(sn),tmp2(sn);
        tmp1[gr] = true;
        dfs4(gr+1,tmp1);
        tmp2[gr] = false;
        dfs4(gr+1,tmp2);
    }
}

int main() {
    cin >> d >> g;
    qn = vector<int>(d);bn = vector<int>(d);
    for (int i=0;i<d;i++) {
        int p,c; cin >> p >> c;
        qn[i] = p;bn[i] = c;
    }

    stb = vector<int>(d);
    st_nob = vector<int>(d);
    for (int i=0;i<d;i++) {
        stb[i] =    (i+1) * qn[i]*100 + bn[i];
        st_nob[i] = (i+1) * (qn[i]-1)*100;
    }
    vector<bool> s(d,false);
    dfs4(0,s);
    cout << mini << endl;
    return 0;
}