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
#include <bits/stdc++.h>
#define debug(x) cerr<<#x<<'='<<x<<endl
#define set0(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
template<typename T> void read(T &x){
	x = 0;char ch = getchar();ll f = 1;
	while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
	while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
bitset<50050> alv[50050];
int n,m,u[100050],v[100050],vis[50050];
char ans[100050];
void solve1(){
	for(int i=1;i<=n;i++){
		alv[i].reset();
		alv[i][i] = 1;
	}
	for(int i=m;i>=1;i--){
		alv[u[i]]|=alv[v[i]];
		alv[v[i]]|=alv[u[i]];
	}
	bitset<50050> cc;
	cc.set();
	for(int i=1;i<=n;i++){
		cc&=alv[i];
	}
	int cho = 0;
	for(int i=1;i<=n;i++) if(cc[i] && !cho)cho = i;
	if(!cho){
		cout<<-1<<endl;
		return;
	}
	vis[cho] = 1;
	for(int i=m;i>=1;i--){
		ans[i] = vis[u[i]]?'^':'v';
		vis[u[i]]|=vis[v[i]];
		vis[v[i]]|=vis[u[i]];
	}
	cout<<(ans+1)<<endl;
}
int cp[50050],cnt[50050];
void solve2(){
	if(n<=2){
		cout<<-1<<endl;
		return;
	}
	for(int i=1;i<=n;i++){
		cp[i] = i;
		cnt[i] = 1;
	}
	for(int i=m;i>=1;i--){
		if(cnt[cp[u[i]]]>cnt[cp[v[i]]]){
			ans[i] = 'v';
			cnt[cp[u[i]]]--;
			cnt[cp[v[i]]]++;
			cp[u[i]] = cp[v[i]];
		}else{
			ans[i] = '^';
			cnt[cp[v[i]]]--;
			cnt[cp[u[i]]]++;
			cp[v[i]] = cp[u[i]];
		}
	}
	cout<<(ans+1)<<endl;
}
int t;
int main() {
	read(n);read(m);read(t);
	for(int i=1;i<=m;i++){
		read(u[i]);
		read(v[i]);
	}
	if(t == 1){
		solve1();
	}
	if(t == 2){
		solve2();
	}
	return 0;
}
