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 ll long long
#define db long double
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define all(a) a.begin(), a.end()

using namespace std;

void solve2(int n, int m) {
	if (n==2) {
		cout << -1;
		return;
	}

	vector<vector<int> > nxt(n);
	vector<pair<int, int> > edges;
	for (int i = 0; i < m; ++i) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		edges.push_back({u, v});
		nxt[u].push_back(v), nxt[v].push_back(u);
	}

	for (int i = 0; i < n; ++i) reverse(nxt[i].begin(), nxt[i].end());

	int a = 0, b = 1;
	if (edges[0].second == 1) b = 2;

	for (int i = 0; i < m; ++i) {
		int u = edges[i].first, v = edges[i].second;
		nxt[u].pop_back(), nxt[v].pop_back();

		if (b == u || b == v) swap(a, b);

		if (u == a) {
			int Q = -1;
			if (nxt[b].size()) Q = nxt[b].back();
			if (Q == a) {
				cout << 'v';
				a = v;
			}
			else cout << '^';
		}
		else if (v == a) {
			int Q = -1;
			if (nxt[b].size()) Q = nxt[b].back();
			if (Q == a) {
				cout << '^';
				a = u;
			}
			else cout << 'v';
		}
		else cout << '^';
	}

}

const int N = 50000;

bitset<N> arr[N];

void solve1(int n, int m) {
	for (int i = 0; i < n; ++i) {
		arr[i][i] = 1;
	}

	vector<pair<int, int> > edges;
	bitset<N> a, b;

	for (int i = 0; i < m; ++i) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		a = arr[u], b = arr[v];
		arr[u] |= b;
		arr[v] |= a;
		edges.push_back({u, v});
	}

	int target = -1;
	for (int i = 0; i < n; ++i) {
		if (arr[i].count() == n) target = i;
	}
	if (target == -1) {
		cout << -1;
		return;
	}

	vector<bool> avail(n, false);
	avail[target] = true;
	vector<char> ans(m);
	for (int i = m - 1; i >= 0; i--) {
		int u = edges[i].first, v = edges[i].second;
		if (!avail[u] && !avail[v]) ans[i] = 'v';
		else if (avail[u] && avail[v]) ans[i] = 'v';
		else if (avail[u]) {
			avail[v] = true;
			ans[i] = '^';
		}
		else{
			avail[u] = true;
			ans[i] = 'v';
		}
	}

	for (int i = 0; i < m; ++i) cout << ans[i];

}


int main(){
#ifdef LOCAL
	freopen("D_input.txt", "r", stdin);
	//freopen("D_output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	int n, m, t;
	cin >> n >> m >> t;

	if (t == 2) solve2(n, m);
	else solve1(n, m);


	
}