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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
    static void Main(string[] args)
    {
            // 入力
            string[] sInp = Console.ReadLine().Trim().Split(' ');
            int iD = int.Parse(sInp[0]);
            int iG = int.Parse(sInp[1]);
            List<ClsSmp> lstSmp = new List<ClsSmp>();
            for (int i = 0; i < iD; i++)
            {
                ClsSmp oneUsr = new ClsSmp();
                sInp = Console.ReadLine().Trim().Split(' ');
                oneUsr.iN = int.Parse(sInp[0]);                 // 問題数
                oneUsr.iB = (i + 1) * 100;                      // 基本点
                oneUsr.iC = int.Parse(sInp[1]);                 // 完全点
                oneUsr.iS = oneUsr.iB * oneUsr.iN + oneUsr.iC; // 小計点(基本点*問題数+完全点)
                lstSmp.Add(oneUsr);
            }

            // 解析
            List<int> lstN = new List<int>();
            lstSmp.Sort(delegate (ClsSmp mca1, ClsSmp mca2) { return (int)(mca2.iB - mca1.iB); });   // 基本点の降順
            for (int i = 0; i < lstSmp.Count; i++)
            {
                fncDo(lstSmp, i, lstN, iG);
            }
            lstSmp.Sort(delegate (ClsSmp mca1, ClsSmp mca2) { return (int)(mca2.iS - mca1.iS); });   // 小計点の降順
            for (int i = 0; i < lstSmp.Count; i++)
            {
                fncDo(lstSmp, i, lstN, iG);
            }
            lstSmp.Sort(delegate (ClsSmp mca1, ClsSmp mca2) { return (int)(mca1.iN - mca2.iN); });   // 問題数の昇順
            for (int i = 0; i < lstSmp.Count; i++)
            {
                fncDo(lstSmp, i, lstN, iG);
            }
            if (lstN.Count > 1) { lstN.Sort(); }

            //出力
            Console.WriteLine("{0}", lstN[0].ToString());

#if DEBUG
        Console.WriteLine("続行するには何かキーを押してください．．．");
        Console.ReadKey();
#endif

    }

        public static void fncDo(List<ClsSmp> lstSmp, int iNext, List<int> lstN, int iG)
        {
            int iC = 0;
            int iN = 0;
            Boolean bOk = false;
            for (int i = iNext; i < lstSmp.Count; i++)
            {
                for (int j = 0; j < lstSmp[i].iN; j++)
                {
                    iN += 1;
                    iC += lstSmp[i].iB;                     // 基本点加算
                    if (iC >= iG) { lstN.Add(iN); bOk = true; break; }
                }
                if (bOk) { break; }
                iC += lstSmp[i].iC;                         // 完全点加算
                if (iC >= iG) { lstN.Add(iN); bOk = true; break; }
            }
        }

        public class ClsSmp
        {
            public int iN { get; set; } = 0;  // 問題数
            public int iB { get; set; } = 0;  // 基本点
            public int iC { get; set; } = 0;  // 完全点
            public int iS { get; set; } = 0;  // 小計点(基本点*問題数+完全点)
        }

}