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;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	const int N=2e5+9;
	int n,m,T,x[N],y[N],ok[N],ans[N],cnt[N];
	bitset<50099> rch[50099],tmp;
	int main(){
		n=read<int>(),m=read<int>(),T=read<int>();
		for(int i=1;i<=m;i++) x[i]=read<int>(),y[i]=read<int>();
		if(T==1){
			for(int i=1;i<=n;i++) rch[i].set(i);
			for(int i=m;i;i--){
				tmp=rch[x[i]]|rch[y[i]];
				rch[y[i]]=rch[x[i]]=tmp;
			}
			tmp=rch[1];
			for(int i=2;i<=n;i++) tmp&=rch[i];

			int pos=-1;
			for(int i=1;i<=n;i++)
				if(tmp[i]){pos=i;break;}
			if(pos==-1) puts("-1"),exit(0);
			
			ok[pos]=1;
			for(int i=m;i;i--) 
				if(ok[x[i]]) ans[i]=1,ok[y[i]]=1;
				else if(ok[y[i]]) ok[x[i]]=1;
		}else{
			for(int i=1;i<=n;i++) ok[i]=i,cnt[i]=1;
			for(int i=m;i;i--){
				if(ok[x[i]]==ok[y[i]]) continue;
			//	cout<<x[i]<<" "<<y[i]<<" "<<ok[x[i]]<<" "<<ok[y[i]]<<endl;
				if(cnt[ok[x[i]]]+1!=n) ans[i]=1,cnt[ok[x[i]]]++,cnt[ok[y[i]]]--,ok[y[i]]=ok[x[i]];
				else if(cnt[ok[y[i]]]+1!=n) cnt[ok[y[i]]]++,cnt[ok[x[i]]]--,ok[x[i]]=ok[y[i]];
				else puts("-1"),exit(0);
			//	cout<<x[i]<<" "<<y[i]<<" "<<ok[x[i]]<<" "<<ok[y[i]]<<endl;
			}
		}
		for(int i=1;i<=m;i++) putchar(ans[i]?'^':'v');
		puts("");
		return 0;
	}
}
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}