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
#include<bits/stdc++.h>
using namespace std;

unsigned long long a[100010][790],tt;
int x[100010],y[100010],n,m,T,nw[50010],Ans,nxt[100010];
char ans[100010];
bool bo[100010],bb[4][4];

int main()
{
	scanf("%d%d%d",&n,&m,&T);
	if (T==1)
	{
		tt=(1ull<<63)-1;
		tt=(tt<<1)+1;
		memset(nw,0,sizeof(nw));
		for (int i=1; i<=m; i++)
		{
			scanf("%d%d",&x[i],&y[i]);
			if (!nw[x[i]]) a[i][x[i]>>6]|=(1ull<<(x[i]&63)); else 
			for (int j=0; j<=(n>>6); j++) a[i][j]|=a[nw[x[i]]][j];
			if (!nw[y[i]]) a[i][y[i]>>6]|=(1ull<<(y[i]&63)); else
			for (int j=0; j<=(n>>6); j++) a[i][j]|=a[nw[y[i]]][j];
			nw[x[i]]=nw[y[i]]=i;
		}
		Ans=0;
		for (int i=1; i<=n; i++)
		{
			bool bo=((a[nw[i]][n>>6]>>1)==(1ull<<(n&63))-1);
			if (n>>6) bo&=(a[nw[i]][n>>6]&1);
			for (int j=0; j<(n>>6); j++) bo&=(a[nw[i]][j]==(tt-(j==0)));
			if (bo) {Ans=i; break;}
		}
		if (!Ans) return puts("-1"),0;
		memset(bo,0,sizeof(bo)),bo[Ans]=1;
		for (int i=m; i; i--)
			if (bo[x[i]]) ans[i]='^',bo[y[i]]=1; else ans[i]='v',bo[x[i]]=bo[y[i]];
		for (int i=1; i<=m; i++) putchar(ans[i]);
		return puts(""),0;
	}
	if (n==2) return puts("-1"),0;
	memset(bb,0,sizeof(bb));
	for (int i=1; i<=m; i++) 
	{
		scanf("%d%d",&x[i],&y[i]);
		if (y[i]<=3) bb[x[i]][y[i]]=1;
	}
	for (int p1=1; p1<=3; p1++)
		for (int p2=p1+1; p2<=3; p2++) if (!bb[p1][p2])
		{
			for (int i=1; i<=m; i++)
				if (x[i]==p1||x[i]==p2) putchar('^'); else putchar('v');
			return puts(""),0;
		}
	int la=-1;
	for (int i=m; i; i--)
		if (y[i]<=3) 
		{
			nxt[i]=la;
			if (x[i]==2) la=1; else
			if (y[i]==2) la=3; else la=2;
		}
	memset(bo,0,sizeof(bo));
	bo[la]=bo[1+(la==1)]=1;
	for (int i=1; i<=m; i++)
		if (y[i]<=3)
		{
			if (x[i]==nxt[i]) putchar('^'),bo[x[i]]=1,bo[y[i]]=0; else
			putchar('v'),bo[y[i]]=1,bo[x[i]]=0;
		} else
		if (x[i]<=3) putchar('^'); else putchar('v');
	return 0;
}