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:
/*input
4 5 2
1 3
2 4
1 2
3 4
2 3
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define eb emplace_back
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define EL cout<<'\n'
#define BS(a,x) binary_search(ALL(a),x)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
	out<<'('<<P.F<<','<<P.S<<')';
	return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
	REP(i,SZ(V)) out<<V[i]<<((i!=SZ(V)-1)?" ":"");
	return out;
}
//}}}
const ll maxn=100005;
const ll maxlg=20;
const ll INF64=1e18;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const ld PI=acos(-1);
const ld eps=1e-9;
//const ll p=880301;
//const ll P=31;

ll mypow(ll a,ll b){
	ll res=1LL;
	while(b){
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}

const int N=50005;

int n,m,t;
int x[maxn],y[maxn];
void solve(){
	if(n==2){
		cout<<"-1\n";
		return;
	}
	int cnt[N]={0};
	int id[N];

	REP(i,n) id[i]=i;
	REP(i,n) cnt[i]=1;

	string ans;
	REP(i,m) cin>>x[i]>>y[i];
	REP(i,m) x[i]--,y[i]--;

	for(int i=m-1;i>=0;i--){
		if(cnt[id[x[i]]]>1){
			ans.pb('v');
			cnt[id[x[i]]]--;
			cnt[id[y[i]]]++;
			id[x[i]]=id[y[i]];
		}
		else {
			ans.pb('^');
			cnt[id[y[i]]]--;
			cnt[id[x[i]]]++;
			id[y[i]]=id[x[i]];
		}
	}
	reverse(ALL(ans));
	cout<<ans<<'\n';
	return;
}


bitset<N> a[N]; 
bitset<N> res;
string ans;

int main(){
	IOS;
	cin>>n>>m>>t;
	if(t==2){
		solve();
		return 0;
	}

	REP(i,n) a[i].reset();
	REP(i,n) a[i][i]=true;

	REP(i,m) cin>>x[i]>>y[i];
	REP(i,m) x[i]--,y[i]--;

	reverse(x,x+m);reverse(y,y+m);

	REP(i,m){
		a[x[i]]|=a[y[i]];
		a[y[i]]|=a[x[i]];
		// REP(j,n) REP(k,n) cout<<a[j][k]<<" \n"[k==n-1];
	}
	REP(i,n) a[0]&=a[i];
	int r=-1;
	REP(i,n) if(a[0][i]){
		r=i;
	}
	// cout<<r<<endl;
	if(r==-1){
		cout<<"-1\n";
		return 0;
	}
	res.reset();
	res[r]=1;
	REP(i,m){
		if(res[x[i]]^res[y[i]]){
			if(res[x[i]]) ans.pb('^');
			else ans.pb('v');

			res[x[i]]=res[y[i]]=true;
		}
		else ans.pb('^');
	}
	reverse(ALL(ans));
	cout<<ans<<'\n';
	return 0;
}