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;

typedef pair <int, int> pii;

const int MAXN = 5e4 + 5;
const int MAXM = 1e5 + 5;
const int SQRT = 450;

int N, M, T;
bitset <MAXN> state[MAXN];
int x[MAXM], y[MAXM];
vector <int> changes[SQRT][SQRT];
vector <pii> todo[MAXM];
bool adj[SQRT][SQRT];
char ans[MAXM];

void load() {
	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]--;
	}
}

void reconstruct(int final) {
	vector <bool> curr(N, false);
	curr[final] = true;
	for (int i = M - 1; i >= 0; i--)
		if (!curr[x[i]]) {
			ans[i] = 'v';
			if (curr[y[i]])
				curr[x[i]] = true;
		}
		else {
			ans[i] = '^';
			if (curr[x[i]])
				curr[y[i]] = true;
		}
	puts(ans);
}

void uniforming() {
	for (int i = 0; i < N; i++)
		state[i].set(i);
	
	for (int i = M - 1; i >= 0; i--)
		state[x[i]] = state[y[i]] = state[x[i]] | state[y[i]];
	
	bitset <MAXN> all;
	all.set();
	for (int i = 0; i < N; i++)
		all &= state[i];
	
	for (int i = 0; i < N; i++)
		if (all.test(i)) {
			reconstruct(i);
			return;
		}
	
	puts("-1");
}

void finish(int u, int v) {
	for (int i = 0; i < M; i++) {
		bool flag = false;
		for (int j = 0; j < 2; j++) {
			if (!changes[u][v].empty() && changes[u][v].back() == i && (u == x[i] || u == y[i])) {
				ans[i] = u == x[i] ? 'v' : '^';
				u = x[i] ^ y[i] ^ u;
				flag = true;
				break;
			}
			swap(u, v);
		}
		if (!flag)
			ans[i] = x[i] == u || x[i] == v ? '^' : 'v';
		for (auto it : todo[i])
			changes[it.first][it.second].pop_back();
	}
	puts(ans);
}

void non_uniforming() {
	map <pii, int> bio;
	for (int i = 0; i < M; i++)
		bio[{x[i], y[i]}] = true;
	
	int sofar = 0;
	for (int i = 0; i < N && sofar <= M; i++)
		for (int j = i + 1; j < N && sofar <= M; j++) {
			if (!bio[{i, j}]) {
				finish(i, j);
				return;
			}
			sofar++;
		}
	
	assert(N < SQRT);	
	
	for (int i = 0; i < N; i++)
		for (int j = 0; j < N; j++)
			adj[i][j] = i != j;

	for (int i = M - 1; i >= 0; i--) {
		adj[x[i]][y[i]] = adj[y[i]][x[i]] = false;
		for (int j = 0; j < N; j++) {
			if (adj[x[i]][j] && !adj[y[i]][j]) 
				todo[i].push_back({y[i], j});
			if (adj[y[i]][j] && !adj[x[i]][j])
				todo[i].push_back({x[i], j});
		}
		for (auto it : todo[i]) {
			adj[it.first][it.second] = adj[it.second][it.first] = true;
			changes[it.first][it.second].push_back(i);
		}
	}
		
	for (int i = 0; i < N; i++)
		for (int j = i + 1; j < N; j++)
			if (adj[i][j]) {
				finish(i, j);
				return;
			}
		
	puts("-1");
}

int main() {
	load();
	if (T == 1)
		uniforming();
	else
		non_uniforming();
	return 0;
}