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;

bitset<50001> bs[50001];
vector<pair<int,int> > e;
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m,t;
	cin>>n>>m>>t;
	for(int i=0,u,v;i<m;i++){
		cin>>u>>v;
		e.emplace_back(u,v);
	}
	string ans;
	ans.assign(m,0);
	if(t==1){
		for(int i=1;i<=n;i++)	bs[i].reset(),bs[i].set(i);
		for(int j=m-1;j>=0;j--){
			int x=e[j].first,y=e[j].second;
			bs[x]|=bs[y];
			bs[y]|=bs[x];
		}
		bs[0].set();
		for(int i=1;i<=n;i++)	bs[0]&=bs[i];
		int r=0;
		for(int i=1;i<=n;i++)
			if(bs[0][i])
				r=i;
		if(r==0)
			cout<<-1<<endl;
		else{
			bs[0].reset();
			bs[0].set(r);
			for(int j=m-1;j>=0;j--){
				int x=e[j].first,y=e[j].second;
				if(bs[0][x])
					ans[j]='^';
				else
					ans[j]='v';
				bs[0][x]=bs[0][x]|bs[0][y];
				bs[0][y]=bs[0][y]|bs[0][x];
			}
			cout<<ans<<endl;
		}
	}
	else{
		if(n<=2)
			cout<<-1<<endl;
		else{
			int state[4]={0,1,2,3};
			for(int i=m-1;i>=0;i--){
				int x=e[i].first,y=e[i].second;
				if(x<=3&&y<=3){
					if(state[x]==state[y])
						ans[i]='^';
					else if(state[x]==state[6-x-y]){
						ans[i]='v';
						state[x]=state[y];
					}
					else{
						ans[i]='^';
						state[y]=state[x];
					}
				}
				else{
					if(x<=3)
						ans[i]='^';
					else
						ans[i]='v';
				}
			}
			cout<<ans<<endl;
		}
	}
}