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"stdafx.h"
#include<vector>
#include<iostream>
#include<fstream>
#include<string>
#include<cassert>
#include<algorithm>
#include<random>
#include<map>
#include<set>
#include <bitset>

using namespace std;

//int mod = 998244353;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int>vi;
typedef vector<ll>vl;
typedef vector<char>vc;
typedef vector<string>vs;
typedef vector<vector<int>>vvi;
typedef vector<vector<char>>vvc;
typedef vector<vector<ll>>vvl;
typedef vector< pair<ll, ll>>vpll;
typedef vector< pair<ld, ld>>vpld;
typedef vector< pair<int, int>>vpi;
typedef pair<ld, ld>pld;
#define mp make_pair
#define pb push_back

int mod=998244353;
int sum(int a, int b)
{
	int c = a + b;
	if (c >= mod)
	{
		c -= mod;
	}
	return c;
}
int dif(int a, int b)
{
	int c = a - b;
	if (c < 0)
	{
		c += mod;
	}
	return c;
}
int mlt(int a, int b)
{
	ll c = a * 1LL * b;
	return c % mod;
}
int ibit(int n, int i)
{
	return ((n >> i) & 1);
}
void outp(vvi &ou)
{
	for (int i = 0; i < ou.size(); i++)
	{
		for (int j = 0; j < ou[i].size(); j++)
		{
			cout << ou[i][j] << ' ';
		}
		cout << '\n';
	}
}
void un(int n, int m, vpi bals)
{
	if (n < 3)
	{
		cout << "-1" << endl;
	}
	else
	{
		vi vals(n), ovals(n), ans(m);
		for (int i = 0; i < n; i++)
		{
			vals[i] = i;
			ovals[i] = 1;
		}
		for (int i = m - 1; i >= 0; i--)
		{
			int x1 = bals[i].first, y1 = bals[i].second;
			int v1 = vals[x1], v2 = vals[y1];
			if (ovals[v1] > 1)
			{
			//	cerr << i << " 1" << endl;
				ans[i] = -1;
				ovals[v1]--;
				ovals[v2]++;
				vals[x1] = v2;
			}
			else
			{
			//	cerr << i << " 2" << endl;
				ans[i] = 1;
				ovals[v2]--;
				ovals[v1]++;
				vals[y1] = v1;
			}
		}
		for (int i = 0; i < m; i++)
		{
			if (ans[i] == -1)
				cout << 'v';
			else
				cout << '^';
		}
		cout << endl;
	}
}
void non_un(int n, int m, vpi bals)
{
	int bits = 50001 / 60 + 1;
	vvl dost(n, vl(bits));
	for (int i = 0; i < n; i++)
	{
		ll x = i % 60;
		dost[i][i/60] += (1LL<<x);
	}
	for (int i = m - 1; i >= 0; i--)
	{
//		cerr << i << ' ' << dost[3][0] << endl;
		int x1 = bals[i].first, y1 = bals[i].second;
		for (int j = 0; j < bits; j++)
		{
			dost[x1][j] |= dost[y1][j];
			dost[y1][j] |= dost[x1][j];
		}
	}
	vl cur = dost[0];
	for (int i = 1; i < n; i++)
		for (int j = 0; j < bits; j++)
			cur[j] &= dost[i][j];

	int fr = -1;
	for (int i = 0; i < bits; i++)
	{
	//	cerr << cur[i] << endl;
		if (cur[i] != 0)
		{
			ll val = cur[i], pos = -1;
			for (int j = 0; j < 61; j++)
			{
				if ((val >> j) & 1)
				{
					pos = j;
					break;
				}
			}
			assert(pos >= 0);
			fr = i * 60 + pos;
	//		cerr << i << ' ' << fr << endl;
			break;
		}
	}
	if (fr == -1)
	{
		cout << "-1" << endl;
	}
	else
	{
		vi used(n), ans(m);
		used[fr] = 1;
		for (int i = m - 1; i >= 0; i--)
		{
			int x1 = bals[i].first, y1 = bals[i].second;
			if (used[x1])
				ans[i] = -1;
			used[x1] |= used[y1];
			used[y1] |= used[x1];
		}
		for (int i = 0; i < m; i++)
		{
			if (ans[i] !=-1)
				cout << 'v';
			else
				cout << '^';
		}
		cout << endl;
	}
}
void solve(istream &cin = std::cin, ostream &cout = std::cout)
{
	int n, m, t;
	cin >> n >> m >> t;
	vpi bals;
	for (int i = 0; i < m; i++)
	{
		int x, y;
		cin >> x >> y;
		bals.pb(mp(x-1, y-1));
	}
	if (t == 1)
	{
		non_un(n, m, bals);
	}
	else
	{
		un(n, m, bals);
	}
}
int main()
{
	solve();
	int n;
	cin >> n;
}