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 N=100005;
int n,m,opt,f[N],cnt[N],a[N],b[N],c[N];
char ans[N];
bitset<N> g[N];
int main(){
	scanf("%d%d%d",&n,&m,&opt);
	for (int i=1;i<=m;i++)scanf("%d%d",&a[i],&b[i]);
	if (opt==1){
		for (int i=1;i<=n;i++)g[i][i]=1;
		for (int i=1;i<=m;i++)g[a[i]]|=g[b[i]],g[b[i]]|=g[a[i]];
		for (int i=1;i<=n;i++)
			if ((int)g[i].count()==n){
				c[i]=1;
				for (int j=m;j;j--){
					if (c[a[j]])ans[j]='^',c[b[j]]=1;
					else ans[j]='v',c[a[j]]=c[b[j]];
				}
				puts(ans+1);
				return 0;
			}
		puts("-1");
	}
	else {
		if (n>=3){
			for (int i=1;i<=n;i++)f[i]=i,cnt[i]++;
			for (int i=m;i;i--){
				if (cnt[f[a[i]]]>1){
					ans[i]='v';
					cnt[f[a[i]]]--;
					cnt[f[b[i]]]++;
					f[a[i]]=f[b[i]];
				}
				else{
					ans[i]='^';
					cnt[f[b[i]]]--;
					cnt[f[a[i]]]++;
					f[b[i]]=f[a[i]];
				}
			}
			puts(ans+1);
		}
		else puts("-1");
	}
}