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>
using namespace std;

int D, G, p[20], c[20], vis[20];

struct T {
    int p, c;
};

int sol() {
    int res = 0, rem = G;
    for(int i = 1 ; i <= D ; ++ i) {
        if(vis[i]) {
            res += p[i];
            rem -= 100 * i * p[i] + c[i];
        }
    }
    for(int i = D ; i >= 1 ; -- i) {
        if(vis[i] == 0) {
            if(rem > 0) {
                if(rem - 100 * i * p[i] >= 0) {
                    // return 1000000000;
                    rem -= 100 * i * p[i] + c[i];
                    res += p[i];                 
                } else {
                    // rem - 100 * i * x >= 0
                    // x <= rem / (100 * i)
                    int x = rem / (100 * i);
                    rem -= 100 * i * x;
                    res += x;
                }
            }
        }
    }
    return rem <= 0 ? res : 1000000000;
}

int main() {
    scanf("%d%d", &D, &G);
    for(int i = 1 ; i <= D ; ++ i) {
        scanf("%d%d", &p[i], &c[i]);
    }
    int ans = 1000000000;
    for(int s = 0 ; s < (1 << D) ; ++ s) {
        for(int i = 1 ; i <= D ; ++ i) vis[i] = ((s >> (i - 1)) & 1);
        ans = min(ans, sol());
        // cout << s << ' ' << sol() << endl;
    }
    printf("%d\n", ans);
}