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 maxn=50005;
const int maxm=100005;
int n,m,T;
int x[maxm],y[maxm];
char s1[maxm],s2[maxm];
bitset<maxn>b[maxn],tot,cur;
set<int>now;
vector<int>diff;
pair<int,int>ans1,ans2;

int main(){
	scanf("%d%d%d",&n,&m,&T);
	for(int i=1;i<=m;i++)scanf("%d%d",&x[i],&y[i]);
	if(T==1){
		for(int i=1;i<=n;i++)b[i][i]=tot[i]=1;
		for(int i=m;i>=1;i--){
			b[x[i]]|=b[y[i]];
			b[y[i]]|=b[x[i]];
		}
		for(int i=1;i<=n;i++)tot&=b[i];
		for(int i=1;i<=n;i++)if(tot[i]){
			cur[i]=1;
			for(int j=m;j>=1;j--){
				if(cur[x[j]])s1[j]='^',cur[y[j]]=1;
				else if(cur[y[j]])s1[j]='v',cur[x[j]]=1;
				else s1[j]='^';
			}
			printf("%s\n",s1+1);
			return 0;
		}
		puts("-1");
	}
	else{
		if(n==2){
			puts("-1");
			return 0;
		}
		for(int i=1;i<=n;i++)now.insert(i);
		for(int i=1;i<=m;i++){
			if(!now.count(x[i])){
				s1[i]=s2[i]='v';
				continue;
			}
			if(!now.count(y[i])){
				s1[i]=s2[i]='^';
				continue;
			}
			if(int(now.size())>3){
				s1[i]=s2[i]='^';
				now.insert(x[i]);
				now.erase(now.find(y[i]));
			}
			else if(ans1.first==0){
				int c;
				for(set<int>::iterator it=now.begin();it!=now.end();it++)
					if((*it)!=x[i]&&(*it)!=y[i])c=(*it);
				diff.push_back(i); 
				ans1=make_pair(x[i],c);
				if(ans1.first>ans1.second)swap(ans1.first,ans1.second);
				s1[i]='^';
				ans2=make_pair(y[i],c);
				if(ans2.first>ans2.second)swap(ans2.first,ans2.second);
				s2[i]='v';
			}
			else{
				if(x[i]==ans1.first&&y[i]==ans1.second){
					for(int j=0;j<int(diff.size());j++)s1[diff[j]]=s2[diff[j]];
					diff.clear();
					int a,b,c;
					if(x[i]==ans2.first)a=x[i],b=y[i],c=ans2.second;
					else if(x[i]==ans2.second)a=x[i],b=y[i],c=ans2.first;
					else if(y[i]==ans2.first)a=y[i],b=x[i],c=ans2.second;
					else a=y[i],b=x[i],c=ans2.first;
					if(a==x[i])s2[i]='^',s1[i]='v';
					else s2[i]='v',s1[i]='^';
					ans1=make_pair(b,c);
					if(ans1.first>ans1.second)swap(ans1.first,ans1.second);
					now.clear();
					now.insert(a);now.insert(b);now.insert(c);
				}
				else if(x[i]==ans2.first&&y[i]==ans2.second){
					for(int j=0;j<int(diff.size());j++)s2[diff[j]]=s1[diff[j]];
					diff.clear();
					int a,b,c;
					if(x[i]==ans1.first)a=x[i],b=y[i],c=ans1.second;
					else if(x[i]==ans1.second)a=x[i],b=y[i],c=ans1.first;
					else if(y[i]==ans1.first)a=y[i],b=x[i],c=ans1.second;
					else a=y[i],b=x[i],c=ans1.first;
					if(a==x[i])s1[i]='^',s2[i]='v';
					else s1[i]='v',s2[i]='^';
					ans2=make_pair(b,c);
					if(ans2.first>ans2.second)swap(ans2.first,ans2.second);
					now.clear();
					now.insert(a);now.insert(b);now.insert(c);
				}
				else{
					if(x[i]==ans1.first||x[i]==ans1.second)s1[i]='^';
					else s1[i]='v';
					if(x[i]==ans2.first||x[i]==ans2.second)s2[i]='^';
					else s2[i]='v';
				}
				diff.push_back(i);
			}
		}
		printf("%s\n",s1+1);
	}
	return 0;
}