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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long long LL;
typedef long double lb;
#define ri register int
const lb PI=3.141592653589793238462643383279;
const ll inf=1000000000000000000ll;
const int N=1000005,M=1;
long long qpow(long long a,long long b)
{
	long long s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		b>>=1;
		s=s*s%M;
	}
	return ans;
}
int n,i,m,ty,x[N],y[N],ans[N],j,flag[N],a[N],e[5],k;
bitset<50005> p[50005],t;
bool f[N];
void dfs(int i)
{
	if(e[1]+e[2]+e[3]<=1)
		return;
	if(i>k)
	{
		for(i=1;i<=m;++i)
			putchar(ans[i]^flag[i]?'v':'^');
		exit(0);
	}
	int tmp[5];
	memcpy(tmp,e,sizeof(tmp));
	e[x[a[i]]]|=e[y[a[i]]];
	e[y[a[i]]]=0;
	ans[a[i]]=0;
	dfs(i+1);
	memcpy(e,tmp,sizeof(tmp));
	e[y[a[i]]]|=e[x[a[i]]];
	e[x[a[i]]]=0;
	ans[a[i]]=1;
	dfs(i+1);
	memcpy(e,tmp,sizeof(tmp));
}
int main()
{
	scanf("%d %d %d",&n,&m,&ty);
	if(ty==2)
	{
		if(n<=2)
		{
			puts("-1");
			return 0;
		}
		for(i=1;i<=m;++i)
		{
			scanf("%d %d",&x[i],&y[i]);
			if(x[i]>y[i])
			{
				flag[i]=1;
				swap(x[i],y[i]);
			}
		}
		e[1]=e[2]=e[3]=1;
		for(i=1;i<=m;++i)
		{
			if((x[i]==1&&y[i]==2)||(x[i]==1&&y[i]==3)||(x[i]==2&&y[i]==3))
				a[++k]=i;
		}
		dfs(1);
		return 0;
	}
	for(i=1;i<=m;++i)
		scanf("%d %d",&x[i],&y[i]);
	for(i=1;i<=n;++i)
		p[i][i]=1;
	for(i=m;i>=1;--i)
		p[x[i]]=p[y[i]]=p[x[i]]|p[y[i]];
	for(i=1;i<=n;++i)
		t[i]=1;
	for(i=1;i<=n;++i)
		t&=p[i];
	if(!t.count())
	{
		puts("-1");
		return 0;
	}
	for(i=1;i<=n;++i)
		if(t[i])
			break;
	f[i]=1;
	for(i=m;i>=1;--i)
		if(f[x[i]])
		{
			f[y[i]]=1;
			ans[i]=1;
		}
		else
		{
			f[x[i]]=f[y[i]];
			ans[i]=0;
		}
	for(i=1;i<=m;++i)
		putchar(ans[i]?'^':'v');
}