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;

using _loop_int = int;
#define REP(i,n) for(_loop_int i=0; i<(_loop_int)(n); i++)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a); i<(_loop_int)(b); i++)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1; i>=(_loop_int)(a); i--)

#define INF 1000000007
#define INFL (1LL << 60)
#define MOD 998244353
#define ALL(v) v.begin(),v.end()
#define pb push_back
#define ll long long int
#define P pair<ll, ll>

ll min(ll a, ll b){return (a<b?a:b);}
ll max(ll a, ll b){return (a>b?a:b);}

int p[100100];
int r[100100];

void init(int n){
  REP(i, n){
    p[i] = i;
    r[i] = 0;
  }
}

int find(int n){
	return (p[n] == n ? n : p[n] = find(p[n]));
}

void unite(int x, int y){
  x = find(x);
  y = find(y);
  if(x == y) return;
  if(r[x] < r[y]){
		p[x] = y;
	}
  else{
    p[y] = x;
    if(r[x] == r[y]) r[x]++;
  }
}

bool same(int x, int y){
  return (find(x) == find(y));
}

int main(){
	int n, m, t;
	cin >> n >> m >> t;
	if(n==2||t==1) {
		cout << -1 << endl;
		return 0;
	}
	P p[m];
	init(n);
	REP(i, m){
		int x, y;
		cin >> x >> y;
		x--;
		y--;
		p[i] = {x, y};
	}
	bool zero[100010] = {};
	REP(i, m){
		if(!zero[p[i].second]){
			unite(p[i].first, p[i].second);
			zero[p[i].second] = true;
		}
	}
	int c = 0;
	REP(i, n) if(same(i, 0)) c++;
	if(c!=n){
		REP(i, m) cout << '^';
		cout << endl;
	}
	else{
		int ind = 0;
		REP(i, m){
			if(p[i].second == n-1){
				ind = p[i].first;
				break;
			}
		}
		REP(i, m){
			if(i != ind) cout << '^';
			else cout << 'v';
		}
		cout << endl;
	}
	return 0;
}