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 <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))

using namespace std;
typedef pair<llint, llint> P;

llint n, m, t;
llint x[100005], y[100005];
bitset<50005> bs[50005];
char ans[100005];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m >> t;
	for(int i = 1; i <= m; i++) cin >> x[i] >> y[i];
	
	if(t == 1){
		for(int i = 1; i <= n; i++) bs[i][i] = 1;
		for(int i = m; i >= 1; i--) bs[x[i]] |= bs[y[i]], bs[y[i]] |= bs[x[i]];
		
		bitset<50005> tmp;
		tmp.set();
		for(int i = 1; i <= n; i++) tmp &= bs[i];
		llint p = 0;
		for(int i = 1; i <= n; i++){
			if(tmp[i]){
				p = i;
				break;
			}
		}
		if(p == 0){
			cout << -1 << endl;
			return 0;
		}
		
		tmp.reset(), tmp[p] = 1;
		for(int i = m; i >= 1; i--){
			if(tmp[x[i]]) ans[i] = '^', tmp[y[i]] = 1;
			else{
				ans[i] = 'v';
				if(tmp[y[i]]) tmp[x[i]] = 1;
			}
		}
	}
	else{
		if(n == 2){
			cout << -1 << endl;
			return 0;
		}
		
		llint p = 0;
		for(int i = m; i >= 1; i--){
			if(y[i] > 3){
				ans[i] = '^';
				continue;
			}
			if(p == 0) ans[i] = '^';
			else{
				if(x[i] == x[p] && y[i] == y[p]) ans[i] = ans[p];
				else{
					if(x[p] != x[i] && y[p] != x[i]) ans[i] = '^';
					else ans[i] = 'v';
				}
			}
			p = i;
		}
	}
	
	for(int i = 1; i <= m; i++) cout << ans[i]; cout << endl;
	
	return 0;
}