Problem:
Score : 100 points

Problem Statement

You are given three integers A, B and C.

Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

Constraints

All values in input are integers.

1 \leq A,B,C \leq 100

Input

Input is given from Standard Input in the following format:

A B C

Output

If there exists an equilateral triangle whose sides have lengths A, B and C, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

There exists an equilateral triangle whose sides have lengths 2, 2 and 2.

Sample Input 2

3 4 5

Sample Output 2

No

There is no equilateral triangle whose sides have lengths 3, 4 and 5.

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

N, Q   = gets.chomp.split.map(&:to_i)
s      = gets.chomp
spells = []
Q.times do
  spells << gets.chomp.split
end
frame = []
N.times do |n|
  frame << [s[n], 1]
end

spells.each_with_index do |spell|
  tmp = Marshal.load(Marshal.dump(frame))
  indexes = []
  frame.each_with_index do |square, i|
    indexes << i if square[0] == (spell[0])
  end

  if spell[1] == "L"
    indexes.each do |n|
      frame[n][1] = 0
      if (n == 0 && spell[1] == "L") || (n == N - 1 && spell[1] == "R")
        tmp[n][1] = 0
      elsif spell[1] == "L"
        frame[n - 1][1] = tmp[n - 1][1] + tmp[n][1]
      elsif spell[1] == "R"
        frame[n + 1][1] = tmp[n + 1][1] + tmp[n][1]
      end
    end
  else
    indexes.reverse.each do |n|
      frame[n][1] = 0
      if (n == 0 && spell[1] == "L") || (n == N - 1 && spell[1] == "R")
        tmp[n][1] = 0
      elsif spell[1] == "L"
        frame[n - 1][1] = tmp[n - 1][1] + tmp[n][1]
      elsif spell[1] == "R"
        frame[n + 1][1] = tmp[n + 1][1] + tmp[n][1]
        if n == 1
        end
      end
    end
  end
end

sum = 0
frame.each { |n| sum += n[1] }
puts sum