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<bits/stdc++.h>
#define randome_generate srand(std::chrono::high_resolution_clock::now().time_since_epoch().count())
using namespace std;
int n,m,t;
bitset<50010> a[50010];
int x[100100],y[100100];
bool rch[50010];
int en[50010];
int cnt[50010];
string s;
void s1(){//subtask 1
	for(int i=1;i<=n;i++){
		a[i].reset();
		rch[i]=false;
	}
	for(int i=1;i<=n;i++)
		a[i].set(i);//自己能到自己 
	for(int i=1;i<=m;i++){//加边 
		a[x[i]]=a[y[i]]=a[x[i]]|a[y[i]];
	}
	for(int i=1;i<=n;i++){//到哪里？ 
		if(a[i].count()==n){//都可到它 
			rch[i]=true;
			for(int j=m;j>=1;j--){
				if(rch[x[j]]){
					rch[y[j]]=rch[x[j]];
					s="^"+s;
				}
				else{
					rch[x[j]]=rch[y[j]];
					s="v"+s;
				}
			}
			cout<<s<<endl;
			return;
		}
	}
	cout<<-1<<endl;
}
void s2(){
	if(n==2){
		cout<<-1<<endl;
		return;
	}
	for(int i=1;i<=n;i++){
		en[i]=i;
		cnt[i]=1;
	}
	for(int i=m;i>=1;i--){
		if(cnt[en[x[i]]]>=2){
			cnt[en[x[i]]]--;
			cnt[en[y[i]]]++;
			en[x[i]]=en[y[i]];
			s="v"+s;
		}
		else{
			cnt[en[y[i]]]--;
			cnt[en[x[i]]]++;
			en[y[i]]=en[x[i]];
			s="^"+s;
		}
	}
	cout<<s<<endl;
}
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>m>>t;
	for(int i=1;i<=m;i++)
		cin>>x[i]>>y[i];
	if(t==1)
		s1();
	else
		s2();
	return 0;
}