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 = 50005;
const int MAXM = 1e5 + 5;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
char s[MAXM]; bitset <MAXN> a[MAXN];
int n, m, t, x[MAXM], y[MAXM], b[MAXN], c[MAXN];
int main() {
	read(n), read(m), read(t);
	if (t == 1) {
		for (int i = 1; i <= n; i++)
			a[i].set(i);
		for (int i = 1; i <= m; i++) {
			read(x[i]), read(y[i]);
			a[x[i]] = a[y[i]] = a[x[i]] | a[y[i]];
		}
		for (int i = 1; i <= n; i++)
			if (a[i].count() == n) {
				static bool vis[MAXN];
				vis[i] = true;
				for (int i = m; i >= 1; i--) {
					if (vis[x[i]]) {
						s[i] = '^';
						vis[y[i]] = true;
					} else if (vis[y[i]]) {
						s[i] = 'v';
						vis[x[i]] = true;
					} else s[i] = '^';
				}
				printf("%s\n", s + 1);
				return 0;
			}
		puts("-1");
	} else {
		if (n == 2) puts("-1");
		else {
			for (int i = 1; i <= n; i++)
				b[i] = i, c[i] = 1;
			for (int i = 1; i <= m; i++)
				read(x[i]), read(y[i]);
			for (int i = m; i >= 1; i--) {
				if (b[x[i]] == b[y[i]]) s[i] = '^';
				else if (c[b[x[i]]] >= 2) {
					c[b[x[i]]]--;
					b[x[i]] = b[y[i]];
					c[b[x[i]]]++;
					s[i] = 'v';
				} else {
					c[b[y[i]]]--;
					b[y[i]] = b[x[i]];
					c[b[y[i]]]++;
					s[i] = '^';
				}
			}
			printf("%s\n", s + 1);
		}
	}
	return 0;
}