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<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<int(n);i++)
#define N 200000
int n, m, t;
int a[N];
int b[N];
char c[N];
long long d[N];
bool used[N];

void p1(void) {
	int x = (n + 59) / 60;
	int y;
	long long z;
	f(i, x) {
		f(j, n)d[j] = 0;
		y = 60 * i;
		z = 1;
		while (y < n) {
			if (y >= (60 * (i + 1)))break;
			d[y] = z;
			y++;
			z *= 2;
		}
		for (int j = m - 1; j >= 0; j--) {
			d[a[j]] = (d[a[j]] | d[b[j]]);
			d[b[j]] = d[a[j]];
		}
		z = z * 2;
		z--;
		f(j, n)z = z & d[j];
		if (z > 0) {
			y = (60 * i);
			while (z > 0) {
				if (z % 2 == 1)break;
				y++;
				z /= 2;
			}
			f(j, n)used[j] = false;
			used[y] = true;
			for (int j = m - 1; j >= 0; j--) {
				if (used[a[j]])c[j] = '^';
				else c[j] = 'v';
				used[a[j]] = (used[a[j]] | used[b[j]]);
				used[b[j]] = used[a[j]];
			}
			printf("%s\n", c);
			return;
		}
	}
	printf("-1\n");
	return;
}

void p2(void) {
	if (n < 3) {
		printf("-1\n");
		return;
	}
	int x = 0;
	for (int i = m - 1; i >= 0; i--) {
		if (b[i] > 2)c[i] = '^';
		else {
			if ((a[i] == 0) && (b[i] == 1)) {
				if (x == 0)c[i] = '^';
				else c[i] = 'v';
				x = 2;
			}
			if ((a[i] == 0) && (b[i] == 2)) {
				if (x == 0)c[i] = '^';
				else c[i] = 'v';
				x = 1;
			}
			if ((a[i] == 1) && (b[i] == 2)) {
				if (x == 1)c[i] = '^';
				else c[i] = 'v';
				x = 0;
			}
		}
	}
	printf("%s\n", c);


	return;
}



int main() {
	f(i, N) {
		a[i] = 0;
		b[i] = 0;
		c[i] = 0;
	}

	scanf("%d %d %d", &n, &m, &t);
	f(i, m) {
		scanf("%d %d", &a[i], &b[i]);
		a[i]--;
		b[i]--;
	}
	if (t == 1)p1();
	else p2();
	return 0;
}