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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.BitSet;
import java.util.StringTokenizer;

public class Main {
	
	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner();
		PrintWriter out = new PrintWriter(System.out);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int T = sc.nextInt();
		int[] x = new int[M], y = new int[M];
		for (int i = 0; i < M; i++) {
			x[i] = sc.nextInt() - 1;
			y[i] = sc.nextInt() - 1;
		}
		if (T == 1) {
			BitSet[] set = new BitSet[N];
			for (int i = 0; i < N; i++) {
				set[i] = new BitSet(N);
				set[i].set(i);
			}
			int last = -1;
			for (int i = 0; i < M && last == -1; i++) {
				set[x[i]].or(set[y[i]]);
				set[y[i]].or(set[x[i]]);
				if (set[x[i]].cardinality() == N) {
					last = x[i];
				}
			}
			if (last == -1) {
				out.println(-1);
			} else {
				BitSet me = new BitSet();
				char[] ret = new char[M];
				me.set(last);
				for (int i = M - 1; i >= 0; i--) {
					if (me.get(x[i])) {
						ret[i] = '^';
						me.set(y[i]);
					} else {
						ret[i] = 'v';
						if (me.get(y[i])) {
							me.set(x[i]);
						}
					}
				}
				out.println(new String(ret));
			}
		}
		if (T == 2) {
			if (N == 2) {
				out.println(-1);
			} else {
				char[] ret = new char[M];
				int[] relevant = new int[M];
				int relSize = 0;
				
				for (int i = 0; i < M; i++) {
					if (y[i] < 3) {
						relevant[relSize++] = i;
					} else {
						ret[i] = '^';
					}
				}
				
				int mask = 7;
				for (int i = 0; i < relSize; i++) {
					int idx = relevant[i];
					int nextMask = 0;
					if (i + 1 < relSize) {
						nextMask = (1 << x[relevant[i+1]]) | (1 << y[relevant[i+1]]);
					}
					int upMask = mask & ~(1<<y[idx]) | (1<<x[idx]);
					if (upMask != nextMask) {
						ret[idx] = '^';
						mask = upMask;
					} else {
						ret[idx] = 'v';
						mask = mask & ~(1<<x[idx]) | (1<<y[idx]);
					}
				}
				
				out.println(new String(ret));
			}
		}
		
		out.flush();
	}
	
	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;
		
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}