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: Time Limit Exceeded)

#include <stdio.h>
int main(void) {
	int  d, g, p[10], c[10],i,s[10],a[10],j,k,b[10],n[10],sum,t,max,l,rem,x,ans,u;
	scanf("%d", &d);
	scanf("%d", &g);
	for (i = 1; i <= d; i++) {
		scanf("%d", &p[i]);
		scanf("%d", &c[i]);
	}
	for (i = 1; i <= d; i++) {
		s[i] = i * p[i] + c[i];
		a[i] = s[i] / p[i];
		b[i] = a[i];
	}
	
	for (i = 0; i<=d-1; i++) {
		for (j = i + 1; j<=d; j++) {
			if (b[i] < b[j]) {
				k = b[i];
				b[i] = b[j];
				b[j] = k;
			}
		}
	}
	for (i = 1; 1 <= d; i++) {
		for (j = 1; j <= d; j++) {
			if (b[i] == a[j]) n[i] = j;
		}
	}
	sum = 0;
	t = 0;
	for (i = 1; sum <= g; i++) {
		sum = sum + s[n[i]]; t++;
	}
	sum = sum - s[n[t]];
	for (i = 1; i <= t - 1; i++) {
		ans = ans + p[n[i]];
	}
	max = 0;
	for(i=1;i<=d;i++){
		l = 0;
		for (j = 1; j <= t - 1; j++) {
			if (i == n[j]) l++;

			if (l == 0) max = i;
		}
		
	}
	rem = g - sum;
	u = 0;
	for (x = 1; rem>=0; x++) {
		rem = rem - max * 100;
		u++;
	}
	u = u - 1;
	ans = ans + u;

	printf("%d", ans);

	return 0;
	}