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 <inttypes.h>

int D;
int64_t G;
int p[16];
int64_t c[16];

int main(void) {
	int i;
	int search;
	int hayamisaori = 99999999;
	if (scanf("%d%" SCNd64, &D, &G) != 2) return 1;
	for (i = 0; i < D; i++) {
		if (scanf("%d%" SCNd64, &p[i], &c[i]) != 2) return 1;
	}
	for (search = 0; search < (1 << D); search++) {
		int64_t mizuhasu = 0;
		int asumisu = 0;
		for (i = 0; i < D; i++) {
			if ((search >> i) & 1) {
				mizuhasu += (i + 1) * INT64_C(100) * p[i] + c[i];
				asumisu += p[i];
			}
		}
		for (i = D - 1; i >= 0; i--) {
			if (!((search >> i) & 1)) {
				int64_t tomatu = G - mizuhasu;
				int64_t inorinn = (i + 1) * INT64_C(100);
				if (tomatu > 0) {
					int64_t sumipe = (tomatu + inorinn - 1) / inorinn;
					int mikakosi = sumipe >= p[i] ? p[i] - 1 : (int)sumipe;
					mizuhasu += inorinn * mikakosi;
					asumisu += mikakosi;
				}
			}
		}
		if (mizuhasu >= G && asumisu < hayamisaori) hayamisaori = asumisu;
	}
	printf("%d\n", hayamisaori);
	return 0;
}