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;

#define form2(i, a, b) for (int i = (a); i < (b); ++i)
#define ford2(i, a, b) for (int i = (a-1); i >= (b); --i)
#define form(i, n) form2(i, 0, n)
#define ford(i, n) ford2(i, n, 0)

#define chmax(x, v) x = max(x, (v))
#define chmin(x, v) x = min(x, (v))
#define fi first
#define se second
#define all(v) v.begin(), v.end()
#define vsz(v) (int)(v.size())

const long long BIG = 1000000000000000000LL;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

void solve();
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	return 0;
}

const int maxLignes = 50*1000 + 5;
bitset<maxLignes> info[maxLignes];
bool canAc[maxLignes];
int repr[maxLignes];
int occ[maxLignes];
int nbLignes, nbEscaliers, mode;

void solve()
{
	cin >> nbLignes >> nbEscaliers >> mode;
	vector<pii> esc;
	form(i, nbEscaliers) {
		int x,y; cin >> x >> y;
		esc.emplace_back(x-1, y-1);
	}
	reverse(all(esc));

	if (mode == 1) {
		form(i, nbLignes) info[i][i] = true;
		for (auto e : esc) {
			info[e.fi] |= info[e.se];
			info[e.se] |= info[e.fi];
		}
		form(i, nbLignes-1) info[i+1] &= info[i];
		int choix = -1;
		form(i, nbLignes) if (info[nbLignes-1][i]) { choix = i; break; }
		if (choix == -1) { cout << "-1\n"; return; }
		canAc[choix] = true;
		string s;
		for (auto e : esc) {
			if (canAc[e.fi]) { s.push_back('^'); canAc[e.se] = true; }
			else { s.push_back('v'); canAc[e.fi] |= canAc[e.se]; }
		}
		reverse(all(s));
		cout << s << "\n";
	} else {
		if (nbLignes == 2) { cout << "-1\n"; return; }
		form(i, nbLignes) { repr[i] = i; occ[i] = 1; }
		string s;
		for (auto e : esc)
		{
			if (occ[repr[e.fi]] >= 2) {
				s.push_back('v');
				--occ[repr[e.fi]];
				++occ[repr[e.se]];
				repr[e.fi] = repr[e.se];
			} else {
				s.push_back('^');
				++occ[repr[e.fi]];
				--occ[repr[e.se]];
				repr[e.se] = repr[e.fi];
			}
		}
		reverse(all(s));
		cout << s << "\n";
	}
}