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

class Program
{
    static void Main(string[] args)
    {
        solve();
    }
    static Scanner cin;

    static int D;
    static int G;
    static int[] p;
    static int[] c;
    static void input()
    {
        cin = new Scanner();
        D = cin.nextInt();
        G = cin.nextInt();
        p = new int[D + 1];
        c = new int[D + 1];
        for(int i = 1; i <= D; i++)
        {
            p[i] = cin.nextInt();
            c[i] = cin.nextInt();
        }
    }
    static void solve()
    {
        input();
        int[,] dp = new int[D + 1,D * 100 + 1];
        int sumP = 0;
        for(int d = 1; d <= D; d++)
        {
            sumP += p[d];
            
            for(int i = 1; i <= sumP; i++)
            {
                for(int j = 0; j <= p[d] && i - j >= 0; j++)
                {
                    dp[d, i] = Max(dp[d - 1, i], dp[d - 1, i - j] + d * 100 * j +
                       (j == p[d] ? c[d] : 0));
                }
            }
        }
        for(int i = 1; ; i++)
        {
            if(dp[D, i] >= G)
            {
                WriteLine(i);
                return;
            }
        }
    }
    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }
        
        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }

        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for(int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for(int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }

        public double[] ArrayDounble(int N, int add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }
}