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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30;
const LL MAX=1e9+7;

void array_show(int *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%d%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(LL *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%lld%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(vector<int> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%d%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}
void array_show(vector<LL> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%lld%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}

char cs[110000];
int t[110000][5];
vector<int> path[110000];

int main(){
	int n,m,p;
	int i,j,k;
	int a,b,c,d;
	cin>>n>>m>>p;
	if(p==1 || n==2){
		cout<<-1<<endl;
		return 0;
	}
	a=-1;
	for(i=0;i<m;i++){
		cin>>t[i][0]>>t[i][1];
		t[i][0]--,t[i][1]--;
		t[i][2]=path[t[i][0]].size();
		t[i][3]=path[t[i][1]].size();
		path[t[i][0]].push_back(i);
		path[t[i][1]].push_back(i);
		if(t[i][2]>0 || t[i][3]>0)a=t[i][0],b=t[i][1];
	}
	if(a==-1){
		
		for(i=0;i<m;i++){
			if(cs[i]!='^' && cs[i]!='v')cs[i]='^';
		}
		cout<<cs<<endl;
		return 0;
	}
	//cout<<a<<b<<endl;
	for(i=0;i<m;i++){
	//cout<<a<<b<<endl;
		if(!(t[i][0]==a || t[i][1]==b || t[i][0]==b || t[i][1]==a))continue;
		
		auto itr=upper_bound(path[a].begin(),path[a].end(),i);
		if(itr==path[a].end())break;
		c=(*itr);
		itr=upper_bound(path[b].begin(),path[b].end(),i);
		if(itr==path[b].end())break;
		d=(*itr);
		if(c==d){
			if((t[c][0]==a && t[c][1]==b)||(t[c][0]==b && t[c][1]==a)){
		//cout<<"y"<<endl;
		//cout<<t[i][0]<<t[i][1]<<endl;
				if(t[i][0]==a)cs[i]='v',a=t[i][1];
				else if(t[i][1]==a)cs[i]='^',a=t[i][0];
				else if(t[i][0]==b)cs[i]='v',b=t[i][1];
				else if(t[i][1]==b)cs[i]='^',b=t[i][0];
				continue;
			}
		}
		
		if(t[i][0]==a)cs[i]='^';
		if(t[i][1]==a)cs[i]='v';
		if(t[i][0]==b)cs[i]='^';
		if(t[i][1]==b)cs[i]='v';
	}
	for(i=0;i<m;i++){
		if(cs[i]!='^' && cs[i]!='v')cs[i]='^';
	}
	cout<<cs<<endl;
}