Problem: 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).

pic1-small-2acea94b.png

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.

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`.

Examples

Input

4 5 1
1 3
2 4
1 2
3 4
2 3


Output

^^^^^


Input

4 5 2
1 3
2 4
1 2
3 4
2 3


Output

v^^^^


Input

3 1 1
1 2


Output

-1


Input

2 1 2
1 2


Output

-1

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <bitset>
using namespace std;
typedef long long int ll;

const int NMAX=50050;
const int MMAX=1e5+5;

int n,m,t;
int x[MMAX],y[MMAX];
bitset<NMAX> bs[NMAX];
bool ok[NMAX];
int go[NMAX];
int cnt[NMAX];

void solve1(){
	for(int i=0;i<m;i++){
		bs[x[i]]=bs[y[i]]=bs[x[i]]|bs[y[i]];
	}
	for(int i=0;i<n;i++){
		if(bs[i].count()==n){
			ok[i]=1;
			string ans(m,'^');
			for(int i=m-1;i>=0;i--){
				if(ok[y[i]]){
					ans[i]='v';
				}
				ok[x[i]]=ok[y[i]]=ok[x[i]]|ok[y[i]];
			}
			cout << ans << endl;
			return;
		}
	}
	cout << -1 << endl;
}

void solve2(){
	if(n==2){
		cout << -1 << endl;
		return;
	}
	for(int i=0;i<n;i++){
		go[i]=i;
		cnt[i]++;
	}
	string ans(m,'^');
	for(int i=m-1;i>=0;i--){
		if(cnt[go[x[i]]]>=2){
			ans[i]='v';
			cnt[go[x[i]]]--;
			go[x[i]]=go[y[i]];
			cnt[go[y[i]]]++;
		}else{
			cnt[go[y[i]]]--;
			go[y[i]]=go[x[i]];
			cnt[go[x[i]]]++;
		}
	}
	cout << ans << endl;
}

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cin >> n >> m >> t;
	for(int i=0;i<n;i++){
		bs[i][i]=1;
	}
	for(int i=0;i<m;i++){
		cin >> x[i] >> y[i];
		x[i]--; y[i]--;
	}
	if(t==1)solve1();
	else solve2();
}