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>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N,M,T;
int A[101010],B[101010];
bitset<50500> S[50505];
int ok[50505],C[50505];


void uniform() {
	int i,j;
	
	FOR(i,M) S[A[i]]=S[B[i]]=S[A[i]]|S[B[i]];
	FOR(i,N) if(S[i].count()==N) {
		ok[i]=1;
		
		string R(M,'^');
		for(i=M-1;i>=0;i--) {
			if(ok[B[i]]) R[i]='v';
			ok[A[i]]=ok[B[i]]=ok[A[i]]|ok[B[i]];
			
		}
		cout<<R<<endl;
		return;
	}
	cout<<-1<<endl;
}

void nonuniform() {
	int i,j;
	
	if(N==2) return _P("-1\n");
	FOR(i,N) {
		ok[i]=i;
		C[i]++;
	}
	
	string R(M,'^');
	for(i=M-1;i>=0;i--) {
		if(C[ok[A[i]]]>=2) {
			R[i]='v';
			C[ok[A[i]]]--;
			ok[A[i]]=ok[B[i]];
			C[ok[A[i]]]++;
		}
		else {
			R[i]='^';
			C[ok[B[i]]]--;
			ok[B[i]]=ok[A[i]];
			C[ok[B[i]]]++;
		}
	}
	cout<<R<<endl;
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>M>>T;
	FOR(i,N) S[i][i]=1;
	FOR(i,M) {
		cin>>A[i]>>B[i];
		A[i]--;
		B[i]--;
	}
	
	if(T==1) {
		uniform();
	}
	else {
		nonuniform();
	}
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}