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)

/**
 *
 * AtCoder Beginner Contest 104 - Problema C: All Green (https://abc104.contest.atcoder.jp/tasks/abc104_c)
 *
 * Check if a given string satisfies all conditions.
 *
 * Ad-hoc solution: check all conditions, one by one.
 */
#include <iostream>

using namespace std;
using ii = pair<int, int>;

const int MAX { 12 };
const int oo { 2000000000 };

int ps[MAX], cs[MAX];

#define LOG(var, sep) (cerr << #var << " = " << (var) << (sep))
#define LOGM(msg) (cerr << (msg))
#define LOGV(vec, sep) { cerr << #vec << " ="; for (const auto& v : vec) cerr << " " << v; cerr << (sep); }
#define LOGA(arr, N, sep) { cerr << #arr << " ="; for (int i = 0; i < N; ++i) cerr << " " << (arr)[i]; cerr << (sep); }
#define LOGA2(arr, N, M, sep) { cerr << #arr << " =\n"; for (int i = 0; i < N; ++i) { for (int j = 0; j < M; ++j) cerr << (arr)[i][j] << " "; cerr << endl; } cerr << (sep); }

void solve(int D, int G)
{
    int min_problems = oo;

    for (int bonus = 0; bonus < (1 << D); ++bonus)
    {
        int problems = 0, left = G;
        
        for (int i = D; i >= 1 and left > 0; --i)
        {
            if (bonus & (1 << (i - 1)))
            {
                problems += ps[i];
                left -= (ps[i] * i * 100 + cs[i]);
            }
        }

        if (left <= 0)
        {
            min_problems = min(min_problems, problems);
            continue;
        }

        for (int i = D; i >= 1 and left > 0; --i)
        {
            if (bonus & (1 << (i - 1)))
                continue;

            int p = min(ps[i] - 1, (left + (100 * i - 1))/(100 * i));
            int value = p * 100 * i;

            problems += p;
            left -= value;
        }

        if (left <= 0)
            min_problems = min(min_problems, problems);
    }

    cout << min_problems << endl;
}

int main()
{
    ios::sync_with_stdio(false);

    int D, G;
    cin >> D >> G;

    for (int i = 1; i <= D; ++i)
        cin >> ps[i] >> cs[i];

    solve(D, G);

    return 0;
}