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;
#define m_p make_pair
const int M = 100005, N = 50004;

int n, m, tt;
pair<int, int> a[M];
char ans[M];

bitset<N> s[N];

void solv1()
{
    for (int i = 1; i <= n; ++i)
        s[i][i] = 1;
    for (int i = m - 1; i >= 0; --i)
    {
        int x = a[i].first;
        int y = a[i].second;
        s[x] |= s[y];
        s[y] |= s[x];
    }
    bitset<N> t;
    for (int i = 1; i <= n; ++i)
        t[i] = 1;
    for (int i = 1; i <= n; ++i)
        t &= s[i];
    int u = -1;
    for (int i = 1; i <= n; ++i)
    {
        if (t[i] == 1)
        {
            u = i;
            break;
        }
    }
    if (u == -1)
    {
        printf("-1\n");
        return;
    }
    for (int i = 1; i <= n; ++i)
        t[i] = 0;
    t[u] = 1;
    for (int i = m - 1; i >= 0; --i)
    {
        int x = a[i].first;
        int y = a[i].second;
        if (t[x] == 1 && t[y] == 0)
        {
            ans[i] = '^';
            t[y] = 1;
        }
        else if (t[x] == 0 && t[y] == 1)
        {
            ans[i] = 'v';
            t[x] = 1;
        }
        else
            ans[i] = '^';
    }
    printf("%s\n", ans);
}

int u[M];
int q[M];

void solv2()
{
	for (int i = 1; i <= n; ++i)
	{
		u[i] = i;
		q[i] = 1;
	}
	for (int i = m - 1; i >= 0; --i)
	{
		if (u[a[i].first] == u[a[i].second])
		{
			ans[i] = '^';
			continue;
		}
		if (q[u[a[i].first]] == n - 1)
		{
			if (q[u[a[i].second]] == n - 1)
			{
				printf("-1\n");
				return;
			}
			ans[i] = 'v';
			--q[u[a[i].first]];
			++q[u[a[i].second]];
			u[a[i].first] = u[a[i].second];
		}
		else
		{
			ans[i] = '^';
			--q[u[a[i].second]];
			++q[u[a[i].first]];
			u[a[i].second] = u[a[i].first];
		}
	}
	printf("%s\n", ans);
}

int main()
{
	scanf("%d%d%d", &n, &m, &tt);
	for (int i = 0; i < m; ++i)
	{
		scanf("%d%d", &a[i].first, &a[i].second);
	}
	if (tt == 1)
        solv1();
	else
        solv2();
	return 0;
}