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 "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"
#include "ctime"

using namespace std;

constexpr long long int MOD = 1000000007;
//constexpr int MOD = 1000000007;
//constexpr int MOD = 998244353;
//constexpr long long int MOD = 998244353;
constexpr double EPS = 1e-9;

//int N, M, K, H, W, L, R;
long long int N, M, K, H, W, L, R;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N >> M >> K;
	if (K == 1) {
		vector<bitset<50000>>bs(150000);
		vector<vector<int>>v(N, vector<int>(1));
		vector<pair<int, int>>w;
		for (int i = 0; i < N; i++) {
			bs[i].flip(i);
			v[i].back() = i;
		}
		for (int i = N; i < N + M; i++) {
			cin >> L >> R;
			L--, R--;
			w.emplace_back(L, R);
			bs[i] = bs[v[L].back()] | bs[v[R].back()];
			v[L].push_back(i);
			v[R].push_back(i);
		}
		bool flag = false;
		int st = 0;
		for (int i = N; i < N + M; i++) {
			if (bs[i].count() == N) {
				st = i - N;
				flag = true;
				break;
			}
		}
		if (!flag) {
			cout << -1 << endl;
			return 0;
		}
		string ans(M, '^');
		vector<int>use(N);
		use[w[st].first] = 1;
		use[w[st].second] = 1;
		for (int i = st - 1; i >= 0; i--) {
			if (!use[w[i].first])ans[i] = 'v';
			if (use[w[i].first] || use[w[i].second]) {
				use[w[i].first] = use[w[i].second] = 1;
			}
		}
		cout << ans << endl;
	}
	else {
		vector<int>v(N, 1);
		if (N == 2) {
			cout << -1 << endl;
			return 0;
		}
		int num = N;
		vector<pair<int, int>>w(M);
		for (auto &i : w) {
			cin >> i.first >> i.second;
			i.first--, i.second--;
		}
		string ans(M, '^');
		for (int i = 0; i < M; i++) {
			if (num == 3) {
				vector<int>p;
				for (int j = 0; j < N; j++) {
					if (v[j])p.push_back(j);
				}
				vector<int>x;
				for (int j = i; j < M; j++) {
					if (v[w[j].first] && v[w[j].second]) {
						x.push_back(j);
					}
				}
				int cnt = 0;
				for (int j = i; j < M; j++) {
					if (cnt < x.size() && x[cnt] == j) {
						if (cnt+1!=x.size()) {
							if (w[x[cnt]] != w[x[cnt + 1]]) {
								if (w[x[cnt]].second != w[x[cnt + 1]].first&&w[x[cnt]].second != w[x[cnt + 1]].second)ans[j] = 'v';
							}
						}
						cnt++;
					}
					else {
						if (v[w[j].second])ans[j] = 'v';
					}
				}
				break;
			}
			num -= v[w[i].first];
			num -= v[w[i].second];
			v[w[i].first] |= v[w[i].second];
			v[w[i].second] = 0;
			num += v[w[i].first];
			num += v[w[i].second];
		}
		cout << ans << endl;
	}
}