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=1e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,T,x[N],y[N],a[N],cnt[N];
char ans[N];
bitset<50010> b[50010];

void mdf(int x,int y) {
	--cnt[a[x]];a[x]=y;++cnt[a[x]];
}

int main() {
	cin>>n>>m>>T;
	for(int i=1;i<=m;i++) x[i]=gi(),y[i]=gi();
	if(T==1) {
		for(int i=1;i<=n;i++) b[i][i]=1;
		for(int i=m;i;i--) b[x[i]]=b[y[i]]=b[x[i]]|b[y[i]];
		bitset<50010> all;for(int i=1;i<=n;i++) all[i]=1;
		for(int i=1;i<=n;i++) all&=b[i];
		if(!all.count()) cout<<-1,exit(0);
		else {
			int p=0;for(int i=1;i<=n;i++) if(all[i]) { p=i;break; }
			a[p]=1;
			for(int i=m;i;i--) {
				if(a[x[i]]&&!a[y[i]]) ans[i]='^';
				else ans[i]='v'; a[x[i]]=a[y[i]]=(a[x[i]]|a[y[i]]);
			}
		}
	}
	else {
		if(n==2) cout<<-1,exit(0);
		for(int i=1;i<=n;i++) a[i]=i,++cnt[i];
		for(int i=m;i;i--) {
			if(cnt[a[x[i]]]>1) ans[i]='v',mdf(x[i],a[y[i]]);
			else ans[i]='^',mdf(y[i],a[x[i]]);
		}
	}
	for(int i=1;i<=m;i++) putchar(ans[i]);
	return 0;
}