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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int n,m,x[100010],y[100010];
namespace work1{
	std::bitset<50001>s[50001],L[100010];
	void main(){
		for(int i=1;i<=n;++i)s[i][i]=s[0][i]=1;
		for(int i=m;i;--i){
			L[i]=s[x[i]]&~s[y[i]];
			s[y[i]]=(s[x[i]]|=s[y[i]]);
		}
		for(int i=1;i<=n;++i)s[0]&=s[i];
		if(!s[0].any())puts("-1");
		else{
			int w=s[0]._Find_first();
			for(int i=1;i<=m;++i)putchar(L[i][w]?'^':'v');
		}
	}
}
namespace work2{
	char ans[100010];
	int to[50010];
	std::map<int,int>M;
	void main(){
		if(n==2)return(void)puts("-1");
		for(int i=1;i<=n;++i)to[i]=i,M[i]=i;
		for(int i=m;i;--i){
			if(to[x[i]]^to[y[i]]){
				if(M[x[i]]>1)ans[i]='v',--M[x[i]],++M[x[i]=y[i]];
				else ans[i]='^',--M[y[i]],++M[y[i]=x[i]];
			}else ans[i]='v';
		}
		printf("%s",ans+1);
	}
}
int main(){
	n=gi(),m=gi();int t=gi();
	for(int i=1;i<=m;++i)x[i]=gi(),y[i]=gi();
	if(t==1)work1::main();
	else work2::main();
	return 0;
}