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 rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const int INF = 1001001001;
const int MOD = 1000000007;

template <typename T>
void print(const T &v);

int main()
{
    int D, G, p[11], c[11];
    cin >> D >> G;
    for (int i = 0; i < D; i++)
    {
        cin >> p[i] >> c[i];
    }

    int ans = 1e9;
    for (int mask = 0; mask < (1 << D); ++mask)
    {
        int s = 0, num = 0, rest_max = -1;
        for (int i = 0; i < D; i++)
        {
            if (mask >> i & 1)
            {
                s += 100 * (i + 1) * p[i] + c[i];
                num += p[i];
            }
            else
            {
                rest_max = i;
            }
        }
        if (s < G)
        {
            int s1 = 100 * (rest_max + 1);
            int need = (G - s + s1 - 1) / s1;
            if (need >= p[rest_max])
            {
                continue;
            }
            num += need;
        }
        ans = min(ans, num);
    }
    cout << ans << endl;
    return 0;
}

// Use For Debug
template <typename T>
void print(T const &v)
{
    for (int i = 0; i < v.size(); i++)
    {
        if (i)
            cout << " ";
        cout << v[i];
    }
    cout << endl;
}