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;
typedef long long ll;
typedef double D;
typedef pair<int,int> P;
typedef tuple<ll,ll,ll> T;
#define M 1000000007
#define F first
#define S second
#define PB push_back
#define INF 1000000000000000000
int n,m,t,x[100005],y[100005],fl[50005];
char ans[100005];
set<int>g[50005];
vector<P>in[100005];
bitset<50005>bs[50005];
int main(void){
    scanf("%d%d%d",&n,&m,&t);
    for(int i=0;i<m;i++)scanf("%d%d",&x[i],&y[i]),x[i]--,y[i]--,ans[i]='v';
    if(t==1){
		for(int i=0;i<n;i++)bs[i][i]=1;
		for(int i=m-1;i>=0;i--)bs[x[i]]|=bs[y[i]],bs[y[i]]|=bs[x[i]];
		for(int i=1;i<n;i++)bs[0]&=bs[i];
		int p=-1;
		for(int i=0;i<n;i++)if(bs[0][i]==1)p=i;
		if(p==-1){
			printf("-1\n");
			return 0;
		}
		fl[p]=true;
		for(int i=m-1;i>=0;i--){
			if(fl[x[i]]){
				ans[i]='^';
				fl[y[i]]=true;
			}else if(fl[y[i]]){
				ans[i]='v';
				fl[x[i]]=true;
			}
		}
		for(int i=0;i<m;i++)printf("%c",ans[i]);
		printf("\n");
		return 0;
	}
	for(int i=0;i<m;i++){
		for(auto it=g[x[i]].begin();it!=g[x[i]].end();it++){
			if(*it==y[i])continue;
			in[i].PB(P(x[i],*it));
			g[*it].erase(x[i]);
		}
		g[x[i]].clear();
		for(auto it=g[y[i]].begin();it!=g[y[i]].end();it++){
			if(*it==x[i])continue;
			in[i].PB(P(y[i],*it));
			g[*it].erase(y[i]);
		}
		g[y[i]].clear();
		g[x[i]].insert(y[i]);
		g[y[i]].insert(x[i]);
	}
	int a,b;
	bool ok=false;
	for(a=0;a<n;a++){
		for(b=a+1;b<n;b++){
			if(g[a].find(b)==g[a].end())ok=true;
			if(ok)break;
		}
		if(ok)break;
	}
	if(!ok){
		printf("-1\n");
		return 0;
	}
	P cur={a,b};
	for(int i=m-1;i>=0;i--){
		bool ch=false;
		for(int j=0;j<in[i].size();j++){
			if(in[i][j].F>in[i][j].S)swap(in[i][j].F,in[i][j].S);
			if(cur==in[i][j]){
				if(x[i]==cur.F||x[i]==cur.S){
					ans[i]='^';
					if(x[i]==cur.F)cur.F=y[i];
					else cur.S=y[i];
				}else if(y[i]==cur.F||y[i]==cur.S){
					ans[i]='v';
					if(y[i]==cur.F)cur.F=x[i];
					else cur.S=x[i];
				}
				ch=true;
				if(cur.F>cur.S)swap(cur.F,cur.S);
			}
		}
		if(!ch){
			if(cur.F==x[i]||cur.S==x[i])ans[i]='^';
			else ans[i]='v';
		}
	}
	for(int i=0;i<m;i++)printf("%c",ans[i]);
	printf("\n");
}