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>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

#define int64 long long

int main()
{
	int64 dd, gg, min = LLONG_MAX, xx;
	scanf("%lld %lld", &dd, &gg);
	int64 pp[dd], cc[dd], score[dd];
	for(int64 ii = 0;ii < dd;ii++)
	{
		scanf("%lld %lld", &pp[ii], &cc[ii]);
		score[ii] = pp[ii] * (ii + 1) * 100 + cc[ii];
	}
	int64 tmp, kosu;
	for(int64 bit = 0;bit < (1 << dd);bit++)
	{
		tmp = 0;
		kosu = 0;
		for(int64 ii = 0;ii < dd;ii++)
		{
			if(bit & (1 << ii))
			{
				tmp += score[ii];
				kosu += pp[ii];
			}
		}
		if(tmp < gg)
		{
			for(int64 ii = dd - 1;ii >= 0;ii--)
			{
				if(!(bit & (1 << ii)))
				{
					xx = (gg - tmp) / ((ii + 1) * 100);
					if((gg - tmp) % ((ii + 1) * 100))
					{
						xx++;
					}
					if(xx > pp[ii])
					{
						break;
					}
					tmp += xx * (ii + 1) * 100;
					kosu += xx;
					break;
				}
			}
		}
		
		if(min > kosu && tmp >= gg)
		{
			min = kosu;
		}
	}
	printf("%lld\n", min);
	return 0;
}