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;

int main() {

	int D;
	unsigned long long int G;
	cin >> D >> G;
	vector<int> p(D);
	vector<unsigned long long int> c(D);
	for (int i = 0; i < D; i++) {
		cin >> p.at(i) >> c.at(i);
	}

	unsigned long long int num,tmp,ans=G;
	bool flag;

	for (int bit = 0; bit < (1 << D); bit++) {
		tmp = 0;
		num = 0;
		flag = true;

		for (int i = 0; i < D; i++) {

			if (bit & (1 << i)) {
				tmp += 100 * (i + 1)*p.at(i) + c.at(i);
				num += p.at(i);
			}

		}

		for (int i = D-1; i >= 0; i--) {
			if (tmp >= G)break;
			if (bit & (1 << i)) {
				continue;
			}
			if (G - tmp > (p.at(i) - 1) * 100 * (i + 1)) {
				flag = false;
				break;
			}

			if ((G - tmp) % (100 * (i + 1)) == 0) {
				num += (G - tmp) / (100 * (i + 1));
				tmp = G;
			}
			else {
				num += (G - tmp) / (100 * (i + 1)) + 1;
				tmp += 100 * (i + 1)*((G - tmp) / (100 * (i + 1)) + 1);
			}

		}

	//	cout << "bit = " << bit << ", num = " << num << ", tmp = " << tmp << ", flag = "<< flag<< endl;
		if (flag && ans > num)ans = num;

	}
	
	
	cout << ans << endl;
}