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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>

using namespace std;
#define MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N, M, T;
int X[101010], Y[101010];

bitset<50000> poss[50000];
bool activ[50000];

void subtask1()
{
	for (int i = 0; i < N; ++i) poss[i].set(i);
	for (int i = 0; i < M; ++i) {
		int x = X[i], y = Y[i];
		poss[x] = poss[x] | poss[y];
		poss[y] = poss[x];
	}
	int target = -1;
	for (int i = 0; i < N; ++i) {
		if (poss[i].count() == N) {
			target = i;
			break;
		}
	}
	if (target == -1) {
		puts("-1");
		return;
	}
	for (int i = 0; i < N; ++i) activ[i] = false;
	activ[target] = true;
	string ans;
	for (int i = M - 1; i >= 0; --i) {
		int x = X[i], y = Y[i];
		if (activ[x] && !activ[y]) {
			ans.push_back('^');
		} else {
			ans.push_back('v');
		}
		if (activ[x] || activ[y]) activ[x] = activ[y] = true;
	}
	reverse(ans.begin(), ans.end());
	printf("%s\n", ans.c_str());
}

void subtask2()
{
	if (N == 2) {
		puts("-1");
		return;
	}
	vector<int> exts;
	for (int i = 0; i < M; ++i) {
		if (Y[i] <= 2) {
			exts.push_back(3 - X[i] - Y[i]);
		}
	}
	vector<int> nn(exts.size(), -1);
	for (int i = 0; i < exts.size(); ++i) {
		int p = i;
		while (p < exts.size() && exts[p] == exts[i]) ++p;
		if (p == exts.size()) break;
		for (int j = i; j < p; ++j) nn[j] = exts[p];
	}
	string ans; int loc = 0;
	for (int i = 0; i < M; ++i) {
		if (Y[i] <= 2) {
			if (nn[loc] == -1) ans.push_back('^');
			else {
				if (X[i] == nn[loc]) ans.push_back('^');
				else ans.push_back('v');
			}
			++loc;
		} else {
			ans.push_back('^');
		}
	}
	printf("%s\n", ans.c_str());
}

int main()
{
	scanf("%d%d%d", &N, &M, &T);
	for (int i = 0; i < M; ++i) {
		scanf("%d%d", X + i, Y + i);
		--X[i]; --Y[i];
	}
	if (T == 1) subtask1();
	else subtask2();
	return 0;
}