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;

const int N=50000,M=100000;

int n,m,opt,x[M+9],y[M+9];

void into(){
  scanf("%d%d%d",&n,&m,&opt);
  for (int i=1;i<=m;++i)
	scanf("%d%d",&x[i],&y[i]);
}

bitset<N+9>dp[N+9];
int vis[N+9];
int ans0,ans[M+9];

void Get_dp(){
  for (int i=1;i<=n;++i) dp[i][i]=1;
  for (int i=1;i<=m;++i){
	dp[x[i]]|=dp[y[i]];
	dp[y[i]]|=dp[x[i]];
  }
  int p=0;
  for (int i=1;i<=n;++i)
	if (dp[i].count()==n) {p=i;break;}
  if (!p) return;
  ans0=1;
  vis[p]=1;
  for (int i=m;i>=1;--i){
	if (!vis[x[i]]&&vis[y[i]]) ans[i]=1;
	vis[x[i]]=vis[y[i]]|=vis[x[i]];
  }
}

int to[N+9],siz[N+9];

void Get_ans(){
  if (n==2&&m) return;
  ans0=1;
  for (int i=1;i<=n;++i) to[i]=i,siz[i]=1;
  for (int i=m;i>=1;--i){
	int u=x[i],v=y[i];
	ans[i]=0;
	if (siz[to[u]]==n-1) swap(u,v),ans[i]=1;
    --siz[to[v]];
	++siz[to[v]=to[u]];
  }
}

void work(){
  opt==1?Get_dp():Get_ans();
}

void outo(){
  if (!ans0) puts("-1");
  else{
	for (int i=1;i<=m;++i)
	  putchar(ans[i]?'v':'^');
    puts("");
  }
}

int main(){
  into();
  work();
  outo();
  return 0;
}