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<cstdio>
#include<bitset>
#include<vector>
const int MaxN=50010,MaxM=100010,MaxB=450;
int N,M,T,x[MaxM],y[MaxM];
std::bitset<MaxN>reach[MaxN];
bool vis[MaxN];
char ans[MaxM];
void solve1(){
	for(int i=1;i<=N;i++)reach[i][i]=1;
	for(int i=0;i<M;i++)reach[x[i]]=reach[y[i]]|=reach[x[i]];
	for(int i=1;i<=N;i++)if(reach[i].count()==N){
		vis[i]=1;
		for(int j=M;j--;){
			if(vis[x[j]]&&!vis[y[j]])vis[y[j]]=1,ans[j]='^';
			else if(!vis[x[j]]&&vis[y[j]])vis[x[j]]=1,ans[j]='v';
			else ans[j]='^';
		}
		puts(ans);
		return;
	}
	puts("-1");
}
bool pairs[MaxB][MaxB];
std::vector<std::pair<int,int> >flips[MaxM];
void flip(int i,int x,int y){
	if(i>=0)flips[i].push_back(std::make_pair(x,y));
	pairs[x][y]=pairs[y][x]^=1;
}
void solve2(){
	int n=1;while(n<N&&n*(n-1)/2<=M)n++;
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)pairs[i][j]=pairs[j][i]=1;
	for(int i=0;i<M;i++)if(y[i]<=n){
		if(pairs[x[i]][y[i]])flip(i,x[i],y[i]);
		for(int j=1;j<=n;j++)if(j!=x[i]&&j!=y[i]&&pairs[x[i]][j]!=pairs[y[i]][j]){
			pairs[x[i]][j]?flip(i,y[i],j):flip(i,x[i],j);
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)if(pairs[i][j]){
			for(int k=M;k--;){
				for(int l=flips[k].size();l--;)flip(-1,flips[k][l].first,flips[k][l].second);
				ans[k]=y[k]==i||y[k]==j?'v':'^';
				if(!pairs[i][j]){
					if(i==x[k])i=y[k];
					else if(i==y[k])i=x[k];
					else if(j==x[k])j=y[k];
					else if(j==y[k])j=x[k];
				}
			}
			puts(ans);
			return;
		}
	puts("-1");
}
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();
	else solve2();
}