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 <bitset>

const int maxn = 50000+7;
const int maxm = 100000+7;

std::pair<int, int> balancer[maxm];
std::bitset<maxn> b[maxn];
int x[maxn], num[maxn];
int n, m, t;

int main(int argc, char *argv[]) {  
	std::ios::sync_with_stdio(false);
	std::cin.tie(0); 
	std::cin >> n >> m >> t;
	for (int i = 0; i < m; i++) {
		std::cin >> balancer[i].first >> balancer[i].second;
		balancer[i].first--;
		balancer[i].second--;
	}
	if (t == 1) {
		for (int i = 0; i < n; i++) {
			b[i].set(i);
		}
		for (int i = 0; i < m; i++) {
			b[balancer[i].first] = b[balancer[i].second] = b[balancer[i].first] | b[balancer[i].second];
		}
		for (int i = 0; i < n; i++) {
			if ((int)b[i].count() == n) {
				std::bitset<maxn> now;
				now.set(i);
				std::string ans(m, 'v');
				while (m--) {
					if (now[balancer[m].first]) {
						now.set(balancer[m].second);
						ans[m] = '^';
					} else {
						now[balancer[m].first] = now[balancer[m].second];
					}
				}
				std::cout << ans << '\n';
				exit(0);
			}
		}
		std::cout << "-1\n";
		exit(0);
	} else {
		if (n == 2) {
			std::cout << "-1\n";
			exit(0);
		}
		std::string ans(m, '^');
		for (int i = 0; i < n; i++) {
			x[i] = i;
			num[i] = 1;
		}
		for (int i = m - 1; i >= 0; i--) {
			if (num[x[balancer[i].first]] > 1) {
				num[x[balancer[i].first]]--;
				num[x[balancer[i].second]]++;
				x[balancer[i].first] = x[balancer[i].second];
				ans[i] = 'v';
			} else {
				num[x[balancer[i].first]]++;
				num[x[balancer[i].second]]--;
				x[balancer[i].second] = x[balancer[i].first];
			}
		}
		std::cout << ans << '\n';
	}
}