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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp51 {
    class Program {
        static void Main(string[] args) {
            var st = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            int D = st[0];
            int G = st[1];

            int[] p = new int[D];
            int[] c = new int[D];

            for (int i = 0; i < D; i++) {
                var t = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
                p[i] = t[0];
                c[i] = t[1];
            }

            int ans = p.Sum();  //初めはansは考えうる最大の値を入れておく

            int nowSum = 0;
            int nowAns = 0;

            //全てパターン試す
            for (int s = 0; s < (1 << D); s++) {
                nowSum = 0;
                nowAns = 0;

                //コンプリートしたところだけをまずは足していく
                for (int i = D - 1; i >= 0; i--) {
                    //今、何問目？
                    int nowPro = D - i - 1;
                    //i桁目のビットが立っているかどうか
                    if ((s & (1 << i)) != 0) {
                        //i問目をコンプリートした
                        nowSum += p[nowPro] * (nowPro + 1) * 100 + c[nowPro];
                        nowAns += p[nowPro];
                    }
                }

                //コンプリートしていないとこを順次足していく
                for (int i = D - 1; i >= 0; i--) {
                    //今、何問目？
                    int nowPro = i;
                    if ((s & (1 << (D - i - 1))) == 0) {
                        for (int j = 1; j <= p[nowPro] - 1; j++) {
                            if (nowSum >= G) {
                                goto warp;
                            }
                            nowSum += (nowPro + 1) * 100;
                            nowAns++;
                        }
                    }
                }

                warp:

                if (nowSum >= G) {
                    ans = Math.Min(ans, nowAns);
                }
            }

            Console.WriteLine(ans);
        }
    }
}