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>
#include <numeric>
#define REP(i, s, n) for (int i = s; i < n; ++i)
#define rep(i, n) REP(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())
#define SORT_INV(c) sort((c).begin(), (c).end(), greater<int>())
#define IINF INT_MAX
#define LLINF LLONG_MAX
#define DEBUG true
#define LL long long
#define Num 1000000007
// sort(a.begin(), a.end(), std::greater<int>());
using namespace std;

int main()
{
    int d, g;
    vector<pair<int, int>> v;
    vector<int> lim;
    vector<int> use;
    cin >> d >> g;
    rep(i, d)
    {
        pair<int, int> hoge;
        cin >> hoge.first >> hoge.second;
        v.push_back(hoge);
    }
    int ans = INT_MAX;
    rep(i, pow(2, d))
    {
        int count = 0;
        int value = 0;
        rep(j, d)
        {
            if ((i & (1 << j)) > 0)
            {
                count += v[j].first;
                value += v[j].first * 100 * (j + 1) + v[j].second;
            }
        }
        if (g <= value)
        {
            //cout << "0Count " << count << "Value " << value << endl;
            ans = min(ans, count);
            continue;
        }
        int use = -1;
        rep(j, d)
        {
            if ((i & (1 << j)) == 0)
            {
                use = j;
            }
        }
        rep(j, v[use].first)
        {
            value += 100 * (use + 1);
            count++;
            if (value >= g)
            {
                //cout << "1Count " << count << "Value " << value << endl;
                ans = min(ans, count);
                break;
            }
        }
    }
    cout << ans << endl;
    return 0;
}