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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
	int d, g;
	scanf("%d %d", &d, &g);
	int p[11], c[11];
	int i, j;
	for (i = 0; i < d; i++)
		scanf("%d %d", &p[i], &c[i]);
	int max = 1;
	for (i = 0; i < d; i++)
		max *= 2;
	int ans = 1000000009;
	int count, score, q;
	int a[11];
	for (i = 0; i < max; i++)
	{
		count = 0;
		score = 0;
		q = i;
		for (j = 0; j < d; j++)
		{
			a[j] = q % 2;
			q /= 2;
		}
		for (j = 0; j < d; j++)
		{
			if (a[j] == 1)
			{
				score += 100 * (j + 1) * p[j] + c[j];
				count += p[j];
			}
		}
		if (score < g)
		{
			for (j = d - 1; j >= 0; j--)
			{
				if (a[j] == 0)
				{
					if (100 * (j + 1) * p[j] + score < g)
					{
						score += 100 * (j + 1) * p[j];
						count += p[j];
					}
					else
					{
						count += (g - score + 100 * (j + 1) - 1) / (100 * (j + 1));
						score += 100 * (j + 1) * ((g - score + 100 * (j + 1) - 1) / (100 * (j + 1)));
						break;
					}
				}
			}
		}
		if (score >= g && ans > count)
			ans = count;
	}
	printf("%d\n", ans);
	return 0;
}