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 REP(i, n) for(ll i = 0; i < n; i++)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define INF LLONG_MAX
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;

int D;
long long G;
vector<ll> p, c;

int main()
{
    cin >> D >> G;
    p.resize(D); c.resize(D);
    for (int i = 0; i < D; ++i) cin >> p[i] >> c[i];

    ll res = 1<<29;
    for (size_t bit = 0; bit < (1<<D); bit++)
    {
        ll sum = 0;
        ll num = 0;
        for (size_t i = 0; i < D; i++)
        {
            if (bit & (1<<i)) sum += c[i] + p[i] * 100 * (i+1), num += p[i];
        }

        if (sum >= G) res = min(res, num);
        else {
            for (int i = D-1; i >= 0; --i) {
                if (bit & (1<<i)) continue;
                for (int j = 0; j < p[i]; ++j) {
                    if (sum >= G) break;
                    sum += 100 * (i+1);
                    ++num;
                }
            }
            res = min(res, num);
        }
        
    }
    cout << res << endl;
    
    return 0;
}