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 LL long long
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
using namespace std;
inline int read(){
	int v=0,f=1;
	char c=getchar();
	while (c<'0' || c>'9'){
		if (c=='-') f=-1;
		c=getchar();
	}
	while (c>='0' && c<='9'){
		v=v*10+c-'0';
		c=getchar();
	}
	return v*f;
}
const int Maxn=50005;
int n,m,t;
int x[Maxn*2],y[Maxn*2];
bitset<Maxn> cur[Maxn];
int f[Maxn],ans[Maxn*2];
void Solve1(){
	for (int i=1;i<=n;i++){
		cur[i][i]=1;
	}
	for (int i=m-1;i>=0;i--){
		cur[x[i]]|=cur[y[i]];
		cur[y[i]]|=cur[x[i]];
	}
	bitset<Maxn> tmp;
	for(int i=1;i<=n;i++) tmp[i]=1;
	for (int i=1;i<=n;i++) tmp&=cur[i];
	for(int i=1;i<=n;i++){
		if (tmp[i]){
			f[i]=1;
			for (int j=m-1;j>=0;j--){
				if (f[x[j]]){
					ans[j]=0;
					f[y[j]]=1;
				} 
				else if (f[y[j]]){
					ans[j]=1;
					f[x[j]]=1;
				}
			}
			for (int j=1;j<=n;j++){
				assert(f[j]);	
			}
			for (int j=0;j<m;j++){
				if (ans[j]){
					putchar('v');
				}
				else{
					putchar('^');
				}
			}
			puts("");
			exit(0);
		}
	}
	printf("-1\n");
}
void Solve2(){
	if (n==2){
		printf("-1\n");
		return;
	}
	int a=-1,lst=-1,b,c,d;
	vector<pair<int,pair<int,int> > > V;
	for (int i=0;i<m;i++){
		int X=x[i],Y=y[i];
		if (X>3) continue;
		if (Y>3){
			ans[i]=0;
			continue;
		}
		V.pb(mp(i,mp(X,Y)));
	}
	if (!V.empty()){
		for (int i=1;i<V.size();i++){
			int NA=6-V[i-1].second.first-V[i-1].second.second,NB=6-V[i].second.first-V[i].second.second;
			if (NA!=NB){
				if (V[i-1].second.first==NB){
					ans[V[i-1].first]=0;
				}
				else{
					ans[V[i-1].first]=1;
				}
			}
		}
	}
	for (int i=0;i<m;i++){
		if (ans[i]){
			putchar('v');
		}
		else{
			putchar('^');
		}
	}
	printf("\n");
}
int main(){
	scanf("%d %d %d",&n,&m,&t);
	for (int i=0;i<m;i++){
		scanf("%d%d",&x[i],&y[i]);
	}
	if (t==1){
		Solve1();
		return 0;
	}
	Solve2();
	return 0;
}