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>

int main(){
    unsigned long int count = 0;
    unsigned long int d, g;
    scanf("%lu%lu", &d, &g);
    g = g/100;
    unsigned long int p[d+1], c[d+1], a[d+1], f[d+1];
    float r[d+1];
    for(int i=1; i<=d; i++){
        scanf("%lu%lu", &p[i], &c[i]);
        c[i] = c[i]/100;
        a[i] = i*p[i] + c[i];
        r[i] = 1.0 * a[i]/p[i];
        f[i] = 1;
    }
    while(g > 0){
        float max = 0.0;
        int id = -1;
        unsigned long int pt = 0;
        unsigned long int x = 0;
        for(int i=d; i>=1; i--){
            if(f[i] == 1){
                float buf;
                unsigned long int pt2, x2;
                if(g <= i*p[i]){
                    buf = i;
                    pt2 = g;
                    x2 = g/i;
                }else{
                    if(a[i] <= g){
                        buf = r[i];
                        pt2 = a[i];
                        x2 = p[i];
                    }else{
                        buf = 1.0 * g/p[i];
                        pt2 = g;
                        x2 = p[i];
                    }
                }
                if(max <= buf){
                    max = buf;
                    id = i;
                    pt = pt2;
                    x = x2;
                }
            }
        }
        count += x;
        g -= pt;
        f[id] = 0;
    }
    printf("%lu", count);

    return 0;
}