Problem:
Score : 1600 points

Problem Statement

A balancing network is an abstract device built up of N wires, thought of as running from left to right,
and M balancers that connect pairs of wires. The wires are numbered from 1 to N from top to bottom,
and the balancers are numbered from 1 to M from left to right.
Balancer i connects wires x_i and y_i (x_i < y_i).

Each balancer must be in one of two states: up or down.

Consider a token that starts moving to the right along some wire at a point to the left of all balancers.
During the process, the token passes through each balancer exactly once.
Whenever the token encounters balancer i, the following happens:

If the token is moving along wire x_i and balancer i is in the down state, the token moves down to wire y_i and continues moving to the right.

If the token is moving along wire y_i and balancer i is in the up state, the token moves up to wire x_i and continues moving to the right.

Otherwise, the token doesn't change the wire it's moving along.

Let a state of the balancing network be a string of length M, describing the states of all balancers.
The i-th character is ^ if balancer i is in the up state, and v if balancer i is in the down state.

A state of the balancing network is called uniforming if a wire w exists such that, regardless of the starting wire,
the token will always end up at wire w and run to infinity along it. Any other state is called non-uniforming.

You are given an integer T (1 \le T \le 2). Answer the following question:

If T = 1, find any uniforming state of the network or determine that it doesn't exist.

If T = 2, find any non-uniforming state of the network or determine that it doesn't exist.

Note that if you answer just one kind of questions correctly, you can get partial score.

Constraints

2 \leq N \leq 50000

1 \leq M \leq 100000

1 \leq T \leq 2

1 \leq x_i < y_i \leq N

All input values are integers.

Partial Score

800 points will be awarded for passing the testset satisfying T = 1.

800 points will be awarded for passing the testset satisfying T = 2.

Input

Input is given from Standard Input in the following format:

N M T
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print any uniforming state of the given network if T = 1, or any non-uniforming state if T = 2. If the required state doesn't exist, output -1.

Sample Input 1

4 5 1
1 3
2 4
1 2
3 4
2 3

Sample Output 1

^^^^^

This state is uniforming: regardless of the starting wire, the token ends up at wire 1.

Sample Input 2

4 5 2
1 3
2 4
1 2
3 4
2 3

Sample Output 2

v^^^^

This state is non-uniforming: depending on the starting wire, the token might end up at wire 1 or wire 2.

Sample Input 3

3 1 1
1 2

Sample Output 3

-1

A uniforming state doesn't exist.

Sample Input 4

2 1 2
1 2

Sample Output 4

-1

A non-uniforming state doesn't exist.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n, m, t = gets.chomp.split().map(&:to_i)
ans = Array.new(m, 'v')
x = Array.new(m)
y = Array.new(m)

m.times do |i|
  xx, yy = gets.chomp.split().map(&:to_i)
  x[i]= xx
  y[i]= yy
end


def kinitsu(n, m, ans, x, y)
  first = true
  pos = 0
  if n > 10
    ar = []
    10.times{ar.push rand(n)}
  else
    ar = (1..n).to_a
  end

  ar.each do |i|
    current = i
    m.times do |j|
      if current == x[j] && ans[j] == 'v'
        current = y[j]
      elsif current == y[j] && ans[j] == '^'
        current = x[j]
      end
    end
    # p ['pos', pos, current]
    if first
      pos = current
      first = false
    elsif pos != current
      return false
    end
  end
  return true
end

if t==1
  # p kinitsu(n,m,'^^v^v'.split(//),x,y)
  puts '-1'
  exit
end

if n == 2
  puts '-1'
  exit
end

8.times do |j|
  n.times do |i|
    ans = Array.new(m, 'v')
    ans[i] = '^' if rand(2) == 1
  end
  if kinitsu(n,m,ans,x,y)
    # skip
  else
    puts ans.join('')
    exit
  end
end

puts '-1'