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
#include<cstdio>
#include<bitset>
#define RI register int
#define CI const int&
using namespace std;
const int N=50005;
int n,m,tp,x[N<<1],y[N<<1]; char s[N<<1];
namespace Case1
{
	bitset <N> vis[N]; bool c[N];
	inline void solve(void)
	{
		RI i,j; for (i=1;i<=n;++i) vis[i].set(i);
		for (i=1;i<=m;++i) vis[x[i]]=vis[y[i]]=vis[x[i]]|vis[y[i]];
		for (i=1;i<=n;++i) if (vis[i].count()==n)
		{
			for (c[i]=1,j=m;j;--j) if (c[x[j]]) s[j]='^',c[y[j]]=1;
			else if (c[y[j]]) s[j]='v',c[x[j]]=1; else s[j]='^';
			return (void)(puts(s+1));
		}
		puts("-1");
	}
};
namespace Case2
{
	int tar[N],c[N];
	inline void solve(void)
	{
		if (n<=2) return (void)(puts("-1")); RI i;
		for (i=1;i<=n;++i) c[tar[i]=i]=1; for (i=m;i;--i)
		if (c[tar[x[i]]]>=c[tar[y[i]]]) s[i]='v',--c[tar[x[i]]],++c[tar[x[i]]=tar[y[i]]];
		else s[i]='^',--c[tar[y[i]]],++c[tar[y[i]]=tar[x[i]]]; puts(s+1);
	}
};
int main()
{
	RI i; for (scanf("%d%d%d",&n,&m,&tp),i=1;i<=m;++i)
	scanf("%d%d",&x[i],&y[i]); if (tp==1) Case1::solve(); else Case2::solve();
	return 0;
}