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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 100001
#define inf 1000000000000000
using namespace std;
struct title {
    int p,c,score;
}s[10];
int d,g,ans;
int vis[10];
int main() {
    scanf("%d%d",&d,&g);
    for(int i = 0;i < d;i ++) {
        scanf("%d%d",&s[i].p,&s[i].c);
        s[i].score = s[i].p * 100 * (i + 1) + s[i].c;
    }
    int sum = 0;
    while(sum < g) {
        int t = -1,n = 101,maxt = -1,maxn = 0;
        for(int i = 0;i < d;i ++) {
            if(vis[i])continue;
            if(s[i].score >= g - sum) {
                int temp;
                if(s[i].score - s[i].c > g - sum) {
                    temp = (g - sum) / 100 / (i + 1);
                }
                else temp = s[i].p;
                if(temp < n) {
                    n = temp;
                    t = i;
                }
            }
            else if(maxt == -1 || s[i].score * s[maxt].p > maxn * s[i].p) {
                maxt = i;
                maxn = s[i].score;
            }
        }
        if(t != -1) {
            if(g - sum <= s[t].p * 100 * (t + 1))ans += (g - sum) / 100 / (t + 1);
            else ans += s[t].p;
            sum = g;
            vis[t] = 1;
        }
        else {
            sum += maxn;
            ans += s[maxt].p;
            vis[maxt] = 1;
        }
    }
    printf("%d",ans);
}