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: Wrong Answer)

#include <stdio.h>
#include <math.h>
#include <string.h>

int mini(int a, int b) {
	if (a < b) {
		return a;
	}
	else {
		return b;
	}
}

int main() {
	int D, i, p[10], c[10], G, min = 1000, sum, count, complete[10], j, rem, quo;

	scanf("%d %d", &D, &G);
	for (i = 0; i < D; i++) {
		scanf("%d %d", &p[i], &c[i]);
	}

	for (i = 0; i < pow(2, D); i++) {
		sum = 0;
		count = 0;
		memset(complete, 0, sizeof(complete));
		for (j = 0; j < D; j++) {
			if ((i >> j) % 2 == 1) {
				sum = sum + c[j] + (j + 1) * 100 * p[j];
				count = count + p[j];
				complete[j] = 1;
			}
			if (sum < G) {
				for (j = D - 1; j >= 0; j--) {
					if (complete[j] == 0) {
						if (G <= sum + (j + 1) * 100 * (p[j] - 1)) {
							rem = (G - sum) % ((j + 1) * 100);
							quo = (G - sum) / ((j + 1) * 100);
							if (rem == 0) {
								count = count + quo;
								sum = sum + (j + 1) * 100 * quo;
							}
							else {
								count = count + quo + 1;
								sum = sum + (j + 1) * 100 * (quo + 1);
							}

						}
						break;
					}
				}
			}
			if (sum >= G) {
				min = mini(min, count);
			}
		}
	}

	printf("%d", min);

	return 0;
}