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 <bits/stdc++.h>
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define int long long
#define INF 10000000000
using namespace std;

typedef pair<int, int> Pii;
typedef pair<double, int> Pdi;
 
void readint(int N, vector<int> &a);
void readdouble(int N, vector<double> &a);
void readindex(int N, vector<int> &a);
 
int calc_num(int i, int p, int c, int G){
    if(G <= 0) return 0;
    if((i + 1) * 100 * p + c < G) return INF;
    if((i + 1) * 100 * p < G) return p;
    
    if(G % ((i + 1) * 100) == 0) return G / ((i + 1) * 100);
    return  G / ((i + 1) * 100) + 1;
}
 
signed main(){
 
    int D, G;
    cin >> D >> G;
    vector<int> p(D), c(D), sump(D);
 
 
    REP(i, D){
        cin >> p[i] >> c[i];
        sump[i] = (i + 1) * 100 * p[i] + c[i];
    }

    vector<int> is_solved(D);
 
    int ans = INF;
    REP(i, 1 << D){
        int sum_tmp = 0;
        int ans_tmp = 0;
        REP(j, D){
            is_solved[j] = (i >> j) % 2;
            if(is_solved[j] > 0){
                sum_tmp += sump[j];
                ans_tmp += p[j];
            }
        }
        if(sum_tmp < G){
            int Gr = G - sum_tmp;
            int imax = -1;
            int num;
            REP(j, D){
                if(is_solved[j] == 0) imax = j;
            }
            if(Gr % ((imax + 1) * 100) == 0){
                num = Gr / ((imax + 1) * 100);
            }else{
                num = Gr / ((imax + 1) * 100) + 1;
            }
            if(num >= p[imax]) continue;
            ans_tmp += num;
        }

        if(ans_tmp < ans) ans = ans_tmp;
    }
 
 
    cout << ans;
 
 
    return 0;
}
 
 
 
 
 
 
 
 
void readint(int N, vector<int> &a){
    string s;
    for(int i = 0; i < N - 1; i++){
        getline(cin, s, ' ');
        a[i] = atoi(s.c_str());
    }
    getline(cin, s, '\n');
    a[N - 1] = atoi(s.c_str());
}
void readdouble(int N, vector<double> &a){
    string s;
    for(int i = 0; i < N - 1; i++){
        getline(cin, s, ' ');
        a[i] = atof(s.c_str());
    }
    getline(cin, s, '\n');
    a[N - 1] = atof(s.c_str());
}
void readindex(int N, vector<int> &a){
    string s;
    for(int i = 0; i < N - 1; i++){
        getline(cin, s, ' ');
        a[i] = atoi(s.c_str()) - 1;
    }
    getline(cin, s, '\n');
    a[N - 1] = atoi(s.c_str()) - 1;
}