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<math.h>
#include<algorithm>
#include<stdint.h>
#include<vector>
#include<deque>
#include<string>
#include<time.h>
#include<list>
#include<random>
#include<bitset>
using namespace std;

using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;



int main() {
    ll d, g;
    cin >> d >> g;
    vector<ll> p(d);
    vector<ll> c(d);
    for (int i = 0; i < d; ++i) {
        cin >> p[i];
        cin >> c[i];
    }
    ll res = 999999999999;
    vector<bool> f(d);
    for (ull i = 0; ((i >> d) & 1) != 1; ++i) {
        ll sum = 0;
        ll mon = 0;
        for (ull j = 0; j < d; ++j) f[j] = (((i >> j) & 1) == 1);
        for (int j = 0; j < d; ++j) {
            if (f[j]) {
                sum += 100 * (j + 1) * p[j];
                sum += c[j];
                mon += p[j];
            }
        }
        for (int j = d - 1; j >= 0 && sum < g; --j) {
            if (f[j]) continue;
            ll point = (100 * (j + 1));
            ll needmon = (g - sum + point - 1) / point;
            if (needmon >= p[j]) {
                sum += point * (p[j] - 1);
                mon += p[j] - 1;
            }
            else {
                sum += point * needmon;
                mon += needmon;
                break;
            }
        }
        if(sum >= g) res = min(res, mon);
    }
    cout << res << endl;
    return 0;
}