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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

// 総数を1000000007（素数）で割った余り
const long long mod = 1e9 + 7;

using ll = long long;
using pii  = pair<int, int>;
using pll = pair<ll, ll>;
#define ull unsigned long long
#define ld long double
#define vi vector<int>
#define vll vector<ll>
#define vc vector<char>
#define vs vector<string>
#define vpii vector<pii>
#define vpll vector<pll>

#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; i++)
#define rep1(i, n) for (int i = 1, i##_len = (n); i <= i##_len; i++)
#define repr(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rep1r(i, n) for (int i = ((int)(n)); i >= 1; i--)

#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define RSORT(x) sort(rall(x));
#define pb push_back
#define mp make_pair

#define INF (1e9)
#define PI (acos(-1))
#define EPS (1e-7)

ull gcd(ull a, ull b) { return b ? gcd(b, a % b) : a; }
ull lcm(ull a, ull b) { return a / gcd(a, b) * b; }

int main(){
    int d, g;
    cin >> d >> g;
    vi p(d), c(d);
    rep(i, d) cin >> p[i] >> c[i];
    g /= 100;
    rep(i, d) c[i] /= 100;
    int ans = INT32_MAX;
    rep(i, 1<<d) {
        int sum = 0, cnt = 0;
        repr(j, d) {
            if (i&(1<<j)) {
                int add = (j+1) * p[j];
                if (sum + add >= g) {
                    int addc = ((g - sum) + ((j+1) - 1)) / (j+1);
                    cnt += addc;
                    sum += addc * (j+1);
                    break;
                }
                sum += add + c[j];
                cnt += p[j];
            }
        }
        if (sum >= g) ans = min(ans, cnt);
    }
    cout << ans << endl;
    return 0;
}