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>
#define PB push_back
#define MP std::make_pair
#define FI first
#define SE second
#define M 1000005
#define INF 1000000000
int n,m,t,X[M],Y[M],num[M],ed[M];
std::bitset <50005> Bitset[50005];
bool flag[M];
char ans[M];
int read(){
	char c=getchar();int ans=0;
	while (c<'0'||c>'9') c=getchar();
	while (c>='0'&&c<='9') ans=ans*10+c-'0',c=getchar();
	return ans;
}
void Write(int x){
	if (x<10) putchar(x+'0');
	else Write(x/10),putchar(x%10+'0');
	return;
}
int main(){
	n=read(),m=read(),t=read();
	for (int i=1;i<=m;i++) X[i]=read(),Y[i]=read();
	if (t==1){
		for (int i=1;i<=n;i++) Bitset[i][i]=Bitset[0][i]=1;
		for (int i=m;i;i--)
			Bitset[X[i]]|=Bitset[Y[i]],Bitset[Y[i]]=Bitset[X[i]];
		for (int i=1;i<=n;i++) Bitset[0]&=Bitset[i];
		if (Bitset[0].count()>0){
			int pos=-1;
			for (int i=1;i<=n;i++)
				if (Bitset[0][i]) pos=i;
			flag[pos]=1;
			for (int i=m;i;i--)
				if (flag[X[i]]==flag[Y[i]]) ans[i]='v';
				else {
					if (flag[X[i]]) ans[i]='^',flag[Y[i]]=1;
					else ans[i]='v',flag[X[i]]=1;
				}
			for (int i=1;i<=m;i++) putchar(ans[i]);
		}
		else printf("-1");
	}
	else {
		if (n==2) printf("-1");
		else {
			for (int i=1;i<=n;i++) ed[i]=i,num[i]=1;
			for (int i=m;i;i--)
				if (ed[X[i]]==ed[Y[i]]) ans[i]='v';
				else {
					if (num[ed[X[i]]]<num[ed[Y[i]]]) num[ed[X[i]]]++,num[ed[Y[i]]]--,ed[Y[i]]=ed[X[i]],ans[i]='^';
					else num[ed[Y[i]]]++,num[ed[X[i]]]--,ed[X[i]]=ed[Y[i]],ans[i]='v';
				}
			for (int i=1;i<=m;i++) putchar(ans[i]);
		}
	}
	return 0;
}