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=114514;
bitset<50005> v[50005];
int x[N],y[N],vis[N],fin[N];
char s[N];
signed main(){
	int n,m,T;
	cin>>n>>m>>T;
	if(T==1){
		for(int i=1;i<=n;++i)
			v[i][i]=1;
		for(int i=1;i<=m;++i){
			cin>>x[i]>>y[i];
			v[x[i]]|=v[y[i]];
			v[y[i]]=v[x[i]];
		}
		for(int i=1;i<=n;++i)
			if(v[i].count()==n){
				vis[i]=1;
				for(int j=m;j>=1;--j)
					if(vis[x[j]]) s[j]='^',vis[y[j]]=1;
					else if(vis[y[j]]) s[j]='v',vis[x[j]]=1;
					else s[j]='v';
				for(int j=1;j<=m;++j)
					cout<<s[j];
				return 0;
			}
		puts("-1"); 
	}
	else{
		if(n<=2) return puts("-1"),0;
		for(int i=1;i<=m;++i)
			cin>>x[i]>>y[i];
		for(int i=1;i<=n;++i)
			vis[i]=1,fin[i]=i;
		for(int i=m;i>=1;--i)
			if(vis[fin[x[i]]]>=vis[fin[y[i]]]) --vis[fin[x[i]]],++vis[fin[y[i]]],fin[x[i]]=fin[y[i]],s[i]='v';
			else --vis[fin[y[i]]],++vis[fin[x[i]]],fin[y[i]]=fin[x[i]],s[i]='^';
		for(int i=1;i<=m;++i)
			cout<<s[i];
	}
	return 0;
}