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;

using ll = long long;

constexpr char endl('\n');

#define FOR(i,l,r) for(size_t i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define INF 100000000000ULL

size_t d, g;

vector<size_t> p;
vector<size_t> c;

size_t dfs(size_t l, size_t n, size_t r, size_t res) {
    if (l == d) {
        if (res >= g) {
            return n;
        }
        else {
            size_t res1 = res + (p.at(r) - 1) * (r + 1);
            if (res1 >= g) {
                size_t i = p.at(r);
                while (res1 >= g) {
                    i--;
                    res1 -= r + 1;
                }
                return n + i;
            }
            else {
                return INF;
            }
        }
    }
    else {
        size_t res1 = dfs(l + 1, n, l, res);
        size_t res2 = dfs(l + 1, n + p.at(l), r, res + p.at(l) * (l + 1) + c.at(l));
        return min(res1, res2);
    }
}

int main()
{
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    
    cin >> d >> g;
    g /= 100;
    p.resize(d);
    c.resize(d);

    REP(i, d) {
        cin >> p.at(i) >> c.at(i);
        c.at(i) /= 100;
    }

    cout << dfs(0, 0, 0, 0) << ::endl;

    return 0;
}