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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define modulo 998244353
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000000000000

unsigned long xor128() {
	static unsigned long x=123456789, y=362436069, z=521288629, w=88675123;
	unsigned long t=(x^(x<<11));
	x=y; y=z; z=w;
	return (w=(w^(w>>19))^(t^(t>>8)));
}

int main(){
	
	int N,M;
	cin>>N>>M;
	
	int T;
	cin>>T;
	
	if(T==1)return 0;
	
	vector<int> a(M),b(M);
	
	for(int i=0;i<M;i++){
		cin>>a[i]>>b[i];
		a[i]--;
		b[i]--;
	}
	
	if(M>=15){
		for(int cnt = 0;cnt < 100;cnt ++){
			string ans = "";
			for(int i=0;i<M;i++){
				if(xor128()%2==0)ans += '^';
				else ans += 'v';
			}
			set<int> S;
			for(int j=0;j<N;j++){
				int now = j;
				for(int k=0;k<M;k++){
					if(a[k] == now){
						if(ans[k] == 'v')now = b[k];
					}
					if(b[k] == now){
						if(ans[k] == '^')now = a[k];
					}
				}
				S.insert(now);
				if(S.size()!=1)break;
			}
			if(S.size()!=1){
				cout<<ans<<endl;
				return 0;
			}
		}
		cout<<-1<<endl;
		return 0;
	}
	
	if(N>=20){
		cout<<string(M,'v')<<endl;
		return 0;
	}
	
	for(int i=0;i<(1<<M);i++){
		string ans = "";
		for(int j=0;j<M;j++){
			if((1>>j)&1)ans += '^';
			else ans += 'v';
		}
		set<int> S;
		for(int j=0;j<N;j++){
			int now = j;
			for(int k=0;k<M;k++){
				if(a[k] == now){
					if(ans[k] == 'v')now = b[k];
				}
				if(b[k] == now){
					if(ans[k] == '^')now = a[k];
				}
			}
			S.insert(now);
		}
		if(S.size()!=1){
			cout<<ans<<endl;
			return 0;
		}
	}
	
	cout<<-1<<endl;

    return 0;
}