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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,t,b[100010][2];
unsigned long long ok[50050];
char ans[100010];
int cnt[50050],to[50050];
int main()
{
	scanf("%d %d %d",&n,&m,&t);
	for(int i=1;i<=m;++i)scanf("%d %d",&b[i][0],&b[i][1]),--b[i][0],--b[i][1];
	if(t==1)
	{
		for(int i=0;i<n;i+=64)
		{
			memset(ok,0,n<<3);
			for(int j=0;j<64&&i+j<n;++j)ok[i+j]|=1ull<<j;
			for(int j=m;j;--j)ok[b[j][0]]=ok[b[j][1]]=ok[b[j][0]]|ok[b[j][1]];
			unsigned long long s=~(0ull);
		//	for(int j=0;j<n;++j)printf("%llu\n",ok[j]);
			for(int j=0;j<n;++j)s&=ok[j];
		//	printf("%llu\n",s);
			if(s)
			{
				int p;
				for(int j=0;j<=63;++j)if(s>>j&1){p=i+j;break;}
				memset(ok,0,n<<3);
				ok[p]=1;
				for(int j=m;j;--j)
				{
					if(ok[b[j][0]])ans[j]='^';
					else ans[j]='v';
					ok[b[j][0]]=ok[b[j][1]]=ok[b[j][0]]|ok[b[j][1]];
				}
				for(int j=1;j<=m;++j)printf("%c",ans[j]);
				return 0;
			}
		}
		printf("-1");
	}
	else
	{
		if(n==2)return printf("-1"),0;
		for(int i=0;i<n;++i)cnt[i]=1,to[i]=i;
		for(int i=m;i;--i)
		{
			if(to[b[i][0]]==to[b[i][1]])ans[i]='v';
			else if(cnt[to[b[i][0]]]>=2)ans[i]='v',--cnt[to[b[i][0]]],++cnt[to[b[i][1]]],to[b[i][0]]=to[b[i][1]];
			else ans[i]='^',--cnt[to[b[i][1]]],++cnt[to[b[i][0]]],to[b[i][1]]=to[b[i][0]];
		}
		for(int i=1;i<=m;++i)printf("%c",ans[i]);
	}
	return 0;
}