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 <iostream>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <algorithm>
using namespace std;
int n, m, T;
#define Maxn 50010
int x[Maxn << 1], y[Maxn << 1];
bitset<Maxn> b[Maxn];
bool ap[Maxn];
int seq[Maxn << 1];
int to[Maxn], B[Maxn];

inline void rd(int &x)
{
	x = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9')
		ch = getchar();
	while (ch >= '0' && ch <= '9')
	{
		x = x * 10 + ch - '0';
		ch = getchar();
	}
}

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

int main()
{
	rd(n);
	rd(m);
	rd(T);
	for (int i = 1; i <= m; ++i)
	{
		rd(x[i]);
		rd(y[i]);
	}
	if (T == 1)
	{
		for (int i = 1; i <= n; ++i)
			b[i][i] = 1;
		bitset<Maxn> res;
		for (int i = m; i >= 1; --i)
		{
			res = b[x[i]] | b[y[i]];
			b[x[i]] = b[y[i]] = res;
		}
		res = b[1];
		for (int i = 2; i <= n; ++i)
			res &= b[i];
		int flag = -1;
		for (int i = 1; i <= n; ++i)
			if (res[i])
			{
				flag = i;
				break;
			}
		if (flag == -1)
		{
			puts("-1");
			return 0;
		}
		ap[flag] = true;
		for (int i = m; i >= 1; --i)
		{
			if (ap[x[i]] == false && ap[y[i]] == false)
				seq[i] = 0;
			else if (ap[x[i]] == true && ap[y[i]] == true)
				seq[i] = 0;
			else if (ap[x[i]])
				seq[i] = 0;
			else
				seq[i] = 1;
			bool tmp = ap[x[i]] | ap[y[i]];
			ap[x[i]] = ap[y[i]] = tmp;
		}
		for (int i = 1; i <= m; ++i)
			if (seq[i] == 0)
				printf("^");
			else
				printf("v");
	}
	else
	{
		if (n == 2)
		{
			puts("-1");
			return 0;
		}
		for (int i = 1; i <= n; ++i)
		{
			B[i] = i;
			to[i] = 1;
		}
		for (int i = 1; i <= n; ++i)
		{
			B[i] = i;
			to[i] = 1;
		}
		for (int i = m; i >= 1; --i)
		{
			if (B[x[i]] == B[y[i]])
			{
				seq[i] = 0;
				continue;
			}
			if (to[B[x[i]]] >= 2)
			{
				seq[i] = 1;
				to[B[x[i]]]--;
				B[x[i]] = B[y[i]];
				to[B[y[i]]]++;
			}
			else
			{
				seq[i] = 0;
				to[B[y[i]]]--;
				B[y[i]] = B[x[i]];
				to[B[x[i]]]++;
			}
		}
		for (int i = 1; i <= m; ++i)
			if (seq[i] == 0)
				printf("^");
			else
				printf("v");
	}
	return 0;
}