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 itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}

const int N = 50050;
using Bs = bitset<N>;
Bs bss[N];

void solve(int n, int m, vector<pair<int, int>> a) {
	for (int i = 0; i < n; ++i) {
		bss[i].flip(i);
	}
	vector<vector<int>> ed(n);
	vector<int> last(n);
	iota(all(last), 0);
	for (int i = 0; i < m; ++i) {
		int x, y;
		tie(x, y) = a[i];
		bss[x] |= bss[y];
		bss[y] = bss[x];

		int nx = (int)ed.size();
		int ny = (int)ed.size() + 1;
		ed.push_back({last[x], last[y]});
		ed.push_back({last[y], last[x]});
		last[x] = nx;
		last[y] = ny;
	}
	int idx = -1;
	for (int i = 0; i < n; ++i) {
		if ((int)bss[i].count() == n) {
			idx = i;
			break;
		}
	}
	if (idx == -1) {
		puts("-1");
		return;
	}
	idx = last[idx];

	vector<int> pr(ed.size(), -1);
	deque<int> q;
	q.push_back(idx);
	vector<int> dst(ed.size(), -1);
	dst[idx] = 0;
	while (!q.empty()) {
		int v = q.front();
		q.pop_front();
		if (ed[v].empty()) {
			continue;
		}
		for (int i = 0; i < 2; ++i) {
			int u = ed[v][i];
			if (dst[u] == -1) {
				dst[u] = dst[v] + i;
				pr[u] = v;
				if (i) {
					q.push_back(u);
				} else {
					q.push_front(u);
				}
			}
		}
	}
	for (int i = 0; i < n; ++i) {
		assert(dst[i] > -1);
	}
	iota(all(last), 0);
	string res(m, 'v');
	for (int i = 0; i < m; ++i) {
		int v = n + 2 * i;
		if (pr[ed[v][1]] == v) {
			res[i] = '^';
		}
	}
	cout << res << "\n";
}

int main() {
	int n = nxt(), m = nxt(), t = nxt();
	vector<pair<int, int>> a(m);
	for (int i = 0; i < m; ++i) {
		a[i].first = nxt() - 1;
		a[i].second = nxt() - 1;
	}

	if (t == 1) {
		solve(n, m, a);
		return 0;
	}

	if (n == 2) {
		puts("-1");
		return 0;
	}
	vector<int> cnt(n, 1);
	vector<int> where(n);
	iota(all(where), 0);
	string res(m, '?');
	for (int i = m - 1; i >= 0; --i) {
		int x, y;
		tie(x, y) = a[i];
		if (where[x] == where[y]) {
			res[i] = 'v';
			continue;
		}
		if (cnt[where[x]] >= cnt[where[y]]) {
			--cnt[where[x]];
			++cnt[where[y]];
			res[i] = 'v';
			where[x] = where[y];
		} else {
			--cnt[where[y]];
			++cnt[where[x]];
			res[i] = '^';
			where[y] = where[x];
		}
	}
	cout << res << "\n";

	return 0;
}