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 SORT(a) sort(a.begin(), a.end())
#define RSORT(a) sort(a.rbegin(), a.rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define RREP(i, n) for (int i = n - 1; 0 <= i; i--)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define RFOR(i, start, end) for (int i = start - 1; 0 <= i; i--)
#define ALL(a) a.begin(), a.end()
using ll = long long;
using namespace std;
constexpr int INF32       = 1'050'000'000;
constexpr long long INF64 = 4'000'000'000'000'000'000;
constexpr int MOD7        = 1'000'000'007;
constexpr int MOD9        = 1'000'000'009;
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
void print() { std::cout << '\n'; }
template <class H, class... T> void print(H &&head, T &&... args) {
    std::cout << head;
    sizeof...(args) == 0 ? std::cout << "" : std::cout << ' ';
    print(std::forward<T>(args)...);
}
template <class T> void print(std::vector<T> &v) {
    for (int i = 0; i < v.size(); i++) {
        std::cout << v[i];
        i == v.size() - 1 ? std::cout << '\n' : std::cout << ' ';
    }
}
template <class T> void print(std::vector<std::vector<T>> &v) {
    for (int i = 0; i < v.size(); i++) {
        for (int j = 0; j < v[i].size(); j++) {
            std::cout << v[i][j];
            j == v[i].size() - 1 ? std::cout << '\n' : std::cout << ' ';
        }
    }
}
void scan() {}
template <class H, class... T> void scan(H &&head, T &&... args) {
    std::cin >> head;
    scan(std::forward<T>(args)...);
}
template <class T> void scan(std::vector<T> &v) {
    for (auto &&i : v) {
        scan(i);
    }
}

template <class T> T up(T a, T div) { return (a + div - 1) / div; }

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int D, G;
    scan(D, G);
    vector<int> p(D), c(D);
    REP(i, D) scan(p[i], c[i]);

    int ans    = INF32;
    auto check = [&](int sum) { return G <= sum ? true : false; };
    REP(bit, (1 << D)) {
        int sum = 0;
        int cnt = 0;
        REP(i, D) {
            if (bit & (1 << i)) {
                sum += (i + 1) * 100 * p[i] + c[i];
                cnt += p[i];
            }
        }
        if (!check(sum)) {
            RREP(i, D) {
                if (!(bit & (1 << i))) {
                    int diff = G - sum;
                    int add  = min(p[i] - 1, up(diff, ((i + 1) * 100)));
                    sum += 100 * (i + 1) * add;
                    cnt += add;
                    if (check(sum))
                        break;
                }
            }
        }
        if (check(sum))
            chmin(ans, cnt);
    }
    print(ans);

    return 0;
}