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;

const int maxn = 50010;
const int maxm = 100010;

int n, m, T;
vector<int> vec[maxn], G[maxm];
int ex[maxm], ey[maxm];
char str[maxm];

namespace Subtask1 {
	int vis[maxn];
	bitset<maxn> dp[maxm];
	void solve() {
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j + 1 < vec[i].size(); j++) {
				G[vec[i][j+1]].push_back(vec[i][j]);
			}
		}
		for (int i = 1; i <= m; i++) {
			dp[i][ex[i]] = dp[i][ey[i]] = 1;
		}
		int ok = 0;
		for (int i = 1; i <= m; i++) {
			for (int j = 0; j < G[i].size(); j++) {
				int v = G[i][j];
				dp[i] |= dp[v];
			}
			if (dp[i].count() == n) {
				ok = i;
			}
		}
		if (!ok) {
			puts("-1");
			exit(0);
		}
		vis[ex[ok]] = vis[ey[ok]] = ok;
		for (int i = ok - 1; i >= 1; i--) {
			int x = ex[i], y = ey[i];
			if (vis[x] && !vis[y]) {
				vis[y] = i;
				str[i] = '^';
			} else if (vis[y] && !vis[x]) {
				vis[x] = i;
				str[i] = 'v';
			}
		}
		for (int i = 1; i <= m; i++) {
			if (str[i] == '\0') {
				if (vis[ex[i]] < vis[ey[i]]) {
					str[i] = 'v';
				} else str[i] = '^';
			}
		}
		puts(str+1);
	}
}

namespace Subtask2 {
	void solve() {
		if (n == 2) {
			puts("-1");
			exit(0);
		}
		int to[4];
		to[1] = 1, to[2] = 2, to[3] = 3;
		for (int i = m; i >= 1; i--) {
			int x = ex[i], y = ey[i];
			if (x <= 3 && y > 3) {
				str[i] = '^';
			} else if (x > 3 && y <= 3) {
				str[i] = 'v';
			} else if (x > 3 && y > 3) {
				str[i] = '^';
			} else {
				if (to[x] == to[y]) {
					str[i] = '^';
				} else {
					int z = 0;
					for (int j = 1; j <= 3; j++) {
						if (j != x && j != y) {
							z = j;
						}
					}
					if (to[x] == to[z]) {
						str[i] = 'v';
						to[x] = to[y];
					} else {
						str[i] = '^';
						to[y] = to[x];
					}
				}
			}
			puts(str + 1);
		}
	}
}

int main() {
	scanf("%d%d%d", &n, &m, &T);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &ex[i], &ey[i]);
		vec[ex[i]].push_back(i), vec[ey[i]].push_back(i);
	}
	if (T == 1) Subtask1::solve();
	else Subtask2::solve();
	return 0;
}