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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
using System.Numerics;

namespace AtCoder
{
    public class Program
    {
        public static void Main(string[] args)
        {
            //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            Solve();
            //Out.Flush();
            Read();
        }

        static void Solve()
        {
            var (D, G) = Slice2<long>();
            var pc = new (long p, long c)[D];
            for (var i = 0; i < D; ++i)
                pc[i] = Slice2<long>();
            var max = 2L << (int)D;
            var cnt = long.MaxValue;
            //completeする問題を決める
            for (var i = 0; i < max; ++i)
            {
                var num = 0L;
                var point = 0L;
                var used = new bool[D];
                for (long j = 0, t = i; j < D; ++j, t >>= 1)
                {
                    if (0 != (t & 1))
                    {
                        used[j] = true;
                        num += pc[j].p;
                        var qp = (j + 1) * 100;
                        point += pc[j].c + pc[j].p * qp;
                    }
                }
                var rem = G - point;
                for (var j = D - 1; 0 <= j && 0 < rem; --j)
                {
                    var qp = (j + 1) * 100;
                    if (used[j]) continue;
                    if (pc[j].p * qp < rem)
                        num = cnt;
                    else
                        num += (rem + qp - 1) / qp;
                    break;
                }
                cnt = Min(cnt, num);
            }
            WriteLine(cnt);

        }

        static long Mod = 998244353;
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static public (T1 a, T2 b) Slice<T1, T2>()
        {
            var t = ReadLine().Split();
            return (
            (T1)Convert.ChangeType(t[0], typeof(T1)),
            (T2)Convert.ChangeType(t[1], typeof(T2)));
        }
        static public (T a, T b) Slice2<T>()
        {
            var t = ReadLine().Split();
            return (
            (T)Convert.ChangeType(t[0], typeof(T)),
            (T)Convert.ChangeType(t[1], typeof(T)));
        }
        static public (T a, T b, T c) Slice3<T>()
        {
            var t = ReadLine().Split();
            return (
            (T)Convert.ChangeType(t[0], typeof(T)),
            (T)Convert.ChangeType(t[1], typeof(T)),
            (T)Convert.ChangeType(t[2], typeof(T)));
        }
    }
}