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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val (n, m, t) = readLine()!!.split(' ').map(String::toLong)
    val ixyList = mutableListOf<Triple<Int, Int, Int>>()
    (1..m.toInt()).forEach {
        val (x, y) = readLine()!!.split(' ').map(String::toInt)
        ixyList.add(Triple(it, x, y))
    }

    if (t == 2L) {
        return
    }
    for (goal in 1..n.toInt()) {
        val ansArray = Array(m.toInt() + 1) { "v" }
        val numArray = IntArray(n.toInt() + 1) { 1 }
        numArray[0] = 0
        numArray[goal] = 0
        for (i in m.toInt() downTo 1) {
            val (it, tx, ty) = ixyList[i - 1]
            if (numArray[tx] == 0) {
                numArray[ty] = 0
                ansArray[i] = "^"
            } else if (numArray[ty] == 0) {
                numArray[tx] = 0
                ansArray[i] = "v"
            }
        }
        if (numArray.sum() == 0) {
            (1..m).forEach { print(ansArray[it.toInt()]) }
            println()
            return
        }
    }
    println(-1)
}