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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

const int M=50000;
bitset<M> tmp[M],zen;
int n,m,type;
vp a;
string s;

int main(){
	cin>>n>>m>>type;
	a=vp(m);
	s=string(m,'^');
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		a[i]={u-1,v-1};
	}
	if(type==1){
		for(int i=0;i<n;i++) tmp[i][i]=zen[i]=true;
		for(int i=m-1;i>=0;i--){
			int u=a[i].first,v=a[i].second;
			tmp[u]|=tmp[v];
			tmp[v]|=tmp[u];
		}
		int v=-1;
		for(int i=0;i<n;i++) zen&=tmp[i];
		for(int i=0;i<n;i++) if(zen[i]) v=i;
		if(v==-1){
			cout<<-1<<endl;
			return 0;
		}
		vi b(n);
		b[v]++;
		for(int i=m-1;i>=0;i--){
			int u=a[i].first,v=a[i].second;
			if(b[v]) s[i]='v';
			if(b[u]||b[v]) b[u]=b[v]=1;
		}
		cout<<s<<endl;
		return 0;
	}
	if(n==2){
		cout<<-1<<endl;
		return 0;
	}
	vi b;
	for(int i=0;i<m;i++){
		int u=a[i].first,v=a[i].second;
		if(v<3) b.push_back(i);
	}
	if(b.empty()){
		cout<<s<<endl;
		return 0;
	}
	int sz=b.size(),x=0,y=1;
	if(a[b[0]].second==1) y=2;
	for(int i=0;i<sz-1;i++){
		int u=a[b[i]].first,v=a[b[i]].second;
		int u_=a[b[i+1]].first,v_=a[b[i+1]].second;
		if(y==u||y==v) swap(x,y);
		if(a[b[i]]==a[b[i+1]]){
			if(x==v) x=u;
		}
		else{
			if(u==u_||u==v_) s[b[i]]='v';
			if(s[b[i]]=='^'){
				if(x==v) x=u;
			}
			else{
				if(x==u) x=v;
			}
		}
	}
	cout<<s<<endl;
}