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;
const int N=100000+10;
const int M=100000+10;
typedef long long ll;
int n,m,T;
int x[M],y[M],pr[M];
void pri(){
	for(int i=1;i<=m;i++)
		cout<<(pr[i]?'^':'v');
}
namespace uni{
	bitset<N> b[N],f;
	int g[N];
	void work(){
		for(int i=1;i<=n;i++)
			b[i][i]=1;
		for(int i=m;i>=1;i--){
			f=(b[x[i]]|b[y[i]]);
			b[x[i]]=b[y[i]]=f;
		}
		f.set();
		for(int i=1;i<=n;i++)
			f&=b[i];
		int ans=0;
		for(int i=1;i<=n;i++)
			if(f[i]){
				ans=i;
				break;
			}
		if(!ans)
			puts("-1");
		else{
			g[ans]=1;
			for(int i=m;i>=1;i--){
				if(g[x[i]])
					g[y[i]]=1,pr[i]=1;
				else if(g[y[i]])
					g[x[i]]=1;
			}
			pri();
		}
	}
}
namespace nuni{
	int b[N],cnt[N];
	void work(){
		if(n==2)
			puts("-1");
		else{
			for(int i=1;i<=n;i++)
				b[i]=i,cnt[i]=1;
			for(int i=m;i>=1;i--){
				if(cnt[b[y[i]]]>1)
					pr[i]=1,cnt[b[y[i]]]--,cnt[b[x[i]]]++,b[y[i]]=b[x[i]];
				else
					cnt[b[y[i]]]++,cnt[b[x[i]]]--,b[x[i]]=b[y[i]];
			}
			pri();
		}
	}
}
int main(){
	cin>>n>>m>>T;
	for(int i=1;i<=m;i++)
		cin>>x[i]>>y[i];
	if(T==1)
		uni::work();
	else
		nuni::work();
	return 0;
}