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(){
    int count = 0;
    int d, g;
    scanf("%d%d", &d, &g);
    int p[d+1], c[d+1];
    int t[d+1][105501];
    g = g/100;
    for(int i=1; i<=d; i++){
        scanf("%d%d", &p[i], &c[i]);
        c[i] = c[i]/100;
        for(int j=0; j<=105501; j++){
            t[i][j] = -1;
        }
    }
    for(int i=0; i<=105501; i++){
        t[0][i] = -1;
    }
    for(int i=1; i<=d; i++){
        int x = 0;
        for(int j=1; j<=p[i]*i; j++){
            t[i][j] = (j+i-1)/i;
        }
        x = p[i]*i;
        for(int j=1; j<=c[i]; j++){
            t[i][x+j] = p[i];
        }
    }
    int id_i, id_j;
    float max = 0;
    while(g > 0){
        max = 0.0;
        for(int i=1; i<=d; i++){
            for(int j=1; j<=g; j++){
                if(max <= j*1.0/t[i][j]){
                    max = j*1.0/t[i][j];
                    id_i = i;
                    id_j = j;
                }
            }
        }
        //printf("%d %d\n", id_i, id_j);
        count += t[id_i][id_j];
        g -= id_j;
        for(int j=0; j<=id_j; j++){
            t[id_i][j] = -1;
        }
    }
    printf("%d", count);
    /*for(int i=10; i>=0; i--){
        for(int j=d; j>=0; j--){
            printf("%d ", t[j][i]);
        }
        printf("\n");
    }*/

    return 0;
}