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.Console;
using System.Text;
using System.IO;
using static System.Math;
namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static long Mod = (long)1e9 + 7;
        static void Main(string[] args)
        {
            //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            var DG = Sarray();
            var D = DG[0];
            var G = DG[1];
            var Pd = new long[D];
            var cd = new long[D];
            for (var i = 0; i < D; ++i)
            {
                var pc = Sarray();
                Pd[i] = pc[0];
                cd[i] = pc[1];
            }
            var dp = Enumerable.Repeat(-1L, 1001).ToArray();
            dp[0] = 0;
            for (var d = 0; d < D; ++d)
            {
                var next = Enumerable.Repeat(-1L, 1001).ToArray();
                next[0] = 0;
                var pt = 0L;
                for (var i = 1; i <= Pd[d]; ++i)
                {
                    pt = (d + 1) * 100 * i;
                    if (i == Pd[d]) pt += cd[d];
                    for (var j = i; j < dp.Length; ++j)
                    {
                        if (-1 == dp[j - i])
                            break;
                        next[j] = Max(next[j], Max(dp[j], dp[j - i] + pt));
                    }
                }
                dp = next;
            }
            for (var i = 0; i < dp.Length; ++i)
            {
                if (G <= dp[i])
                {
                    WriteLine(i);
                    return;
                }
            }
            //Out.Flush();
        }
    }
}