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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <bitset>

using namespace std;
typedef long long ll;
const int MAXN = 50005, MAXM = 100005;
inline int readint()
{
	int res = 0, f = 1;
	char c = 0;
	while(!isdigit(c))
	{
		c = getchar();
		if(c=='-')
			f = -1;
	}
	while(isdigit(c))
		res = res*10+c-'0', c = getchar();
	return res*f;
}
int n,m,type,x[MAXM],y[MAXM];
bitset<MAXN> f[MAXN];
bool t[MAXN];
char ans[MAXM];
inline void solve1()
{
	for(int i = 1; i<=n; i++)
	    f[i][i] = 1;
	for(int i = m; i>=1; i--)
	    f[x[i]] = f[y[i]] = f[x[i]]|f[y[i]];
	for(int i = 2; i<=n; i++)
	    f[1] &= f[i];
	int now = 0;
	for(int i = 1; i<=n; i++)
        if(f[1][i])
        	now = i;
	if(!now)
	{
    	cout << -1 << endl;
    	return;
	}
	t[now] = true;
	for(int i = m; i>=1; i--)
	{
        if(t[x[i]])
        	ans[i] = '^';
        else
            ans[i] = 'v';
        t[x[i]] = t[y[i]] = t[x[i]]|t[y[i]];
	}
	for(int i = 1; i<=m; i++)
	    cout << ans[i];
	cout << endl;
}
int g[MAXN],h[MAXN];
inline void solve2()
{
	if(n==2&&m>=1)
	{
        cout << -1 << endl;
        return;
	}
	for(int i = 1; i<=n; i++)
		g[i] = 1, h[i] = i;
	for(int i = m; i>=1; i--)
	{
		if(g[h[x[i]]]>g[h[y[i]]])
			g[h[x[i]]]--, g[h[y[i]]]++, h[x[i]] = h[y[i]], ans[i] = 'v';
		else
		    g[h[x[i]]]++, g[h[y[i]]]--, h[y[i]] = h[x[i]], ans[i] = '^';
	}
	for(int i = 1; i<=m; i++)
		cout << ans[i];
	cout << endl;
}

int main()
{
	n = readint(), m = readint(), type = readint();
	for(int i = 1; i<=m; i++)
		x[i] = readint(), y[i] = readint();
	if(type==1)
	    solve1();
	else
	    solve2();
	return 0;
}