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;

int N , M , T , s[100003] , t[100003] , dis[100003];
bitset < 50003 > now[50003];

namespace sub1{
	void work(){
		for(int i = 1 ; i <= N ; ++i) now[i].set(i);
		for(int i = 1 ; i <= M ; ++i) now[s[i]] = now[t[i]] = now[s[i]] | now[t[i]];
		for(int i = 1 ; i <= N ; ++i)
			if(now[i].count() == N){
				dis[i] = 100000;
				for(int j = M ; j ; --j)
					if(dis[s[j]] && !dis[t[j]]) dis[t[j]] = j;
					else if(dis[t[j]] && !dis[s[j]]) dis[s[j]] = j;
				for(int j = 1 ; j <= M ; ++j)
					putchar(dis[s[j]] < dis[t[j]] ? 'v' : '^');
				return;
			}
		puts("-1");
	}
}

int pot[50003] , f[50003]; char str[100003];
namespace sub2{
	void work(){
		if(N == 2){puts("-1"); return;}
		for(int i = 1 ; i <= N ; ++i) pot[f[i] = i] = 1;
		for(int j = M ; j ; --j)
			if(pot[f[s[j]]] < pot[f[t[j]]]){str[j] = '^'; --pot[f[t[j]]]; ++pot[f[t[j]] = f[s[j]]];}
			else{str[j] = 'v'; --pot[f[s[j]]]; ++pot[f[s[j]] = f[t[j]]];}
		printf("%s\n" , str + 1);
	}
}

int main(){
	ios::sync_with_stdio(0); cin >> N >> M >> T;
	for(int i = 1 ; i <= M ; ++i) cin >> s[i] >> t[i];
	if(T == 1) sub1::work(); else sub2::work();
	return 0;
}