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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

const int MAXN = 300005;

struct Edge
{
	int to;
	int w;
	int nxt;
}edge[MAXN << 2];

int n,m,T,st,ed,id;
int cur[MAXN];
int u[MAXN];
int v[MAXN];
int first[MAXN];
int cnt[MAXN];

bool vis[MAXN];
bool ans[MAXN];

bitset<50001> mask[150005];

void addE(int u,int v,int w)
{
	edge[++id] = (Edge){v,w,first[u]};
	first[u] = id;
}

void dfs(int u)
{
	vis[u] = 1;
	for (int i = first[u];i;i = edge[i].nxt)
		if (!vis[edge[i].to])
		{
			if (edge[i].w)
				ans[edge[i].w] = edge[i].to < u;
			dfs(edge[i].to);
		}
}

int main()
{
	scanf("%d%d%d",&n,&m,&T);
	for (int i = 1;i <= m;i++)
		scanf("%d%d",&u[i],&v[i]);
	if (T == 1)
	{
		for (int i = 1;i <= n;i++)
			cur[i] = i,mask[i][i] = 1;
		for (int i = 1;i <= m;i++)
		{
			mask[n + i] = mask[cur[u[i]]] | mask[cur[v[i]]];
			cur[u[i]] = cur[v[i]] = n + i;
			if (mask[n + i].count() == n)
			{
				ed = i;
				st = u[i];
				break;
			}
		}
		if (!st)
		{
			puts("-1");
			return 0;
		}
		for (int i = 1;i <= n;i++)
			cur[i] = i;
		int nd = n;
		for (int i = 1;i <= ed;i++)
		{
			addE(++nd,cur[u[i]],0);
			cur[u[i]] = nd;
			addE(++nd,cur[v[i]],0);
			cur[v[i]] = nd;
			addE(nd - 1,nd,i);
			addE(nd,nd - 1,i);
		}
		memset(vis,0,sizeof(vis));
		dfs(cur[st]);
		for (int i = 1;i <= m;i++)
			putchar(ans[i] ? 'v' : '^');
		putchar('\n');
	}
	else
	{
		if (n == 2)
		{
			puts("-1");
			return 0;
		}
		for (int i = 1;i <= n;i++)
			cnt[cur[i] = i] = 1;
		for (int i = m;i >= 1;i--)
		{
			if (cnt[cur[u[i]]] > 1)
			{
				ans[i] = 1;
				cnt[cur[u[i]]]--;
				cur[u[i]] = cur[v[i]];
				cnt[cur[v[i]]]++;
			}
			else
			{
				ans[i] = 0;
				cnt[cur[v[i]]]--;
				cur[v[i]] = cur[u[i]];
				cnt[cur[u[i]]]++;
			}
		}
		for (int i = 1;i <= m;i++)
			putchar(ans[i] ? 'v' : '^');
		putchar('\n');
	}
	return 0;
}