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<iostream>
#include<cstring>
#include<cstdio>
#include<bitset>
#include<algorithm>
using namespace std;
int n,m,T;
#define Maxn 50010
int x[Maxn<<1],y[Maxn<<1];
bitset<Maxn> b[Maxn];
bool ap[Maxn];
int seq[Maxn<<1];
int to[Maxn],B[Maxn];

inline void rd(int &x){
	x=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9'){
		x=x*10+ch-'0';
		ch=getchar();
	}
}

int main(){
	rd(n);rd(m);rd(T);
	for(int i=1;i<=m;++i){
		rd(x[i]);rd(y[i]);
    }
    if(T==1){
		for(int i=1;i<=n;++i)b[i][i]=1;
		bitset<Maxn> res;
		for(int i=m;i>=1;--i){
			res=b[x[i]]|b[y[i]];
			b[x[i]]=b[y[i]]=res;
		}
		res=b[1];
		for(int i=2;i<=n;++i)res&=b[i];
		int flag=-1;
		for(int i=1;i<=n;++i)
		    if(res[i]){
				flag=i;
				break;
		    } 
		if(flag==-1){
			puts("-1");
			return 0;
		}
		ap[flag]=true;
		for(int i=m;i>=1;--i){
			if(ap[x[i]]==false&&ap[y[i]]==false)seq[i]=0;
			else if(ap[x[i]]==true&&ap[y[i]]==true)seq[i]=0;
		    else if(ap[x[i]])seq[i]=0;
		    else seq[i]=1;
		    bool tmp=ap[x[i]]|ap[y[i]];
		    ap[x[i]]=ap[y[i]]=tmp;
		}
		for(int i=1;i<=m;++i)
		    if(seq[i]==0)printf("^");
		    else printf("v");
    }else{
		if(n==2){
			puts("-1");
			return 0;
		}
		for(int i=1;i<=n;++i){
			B[i]=i;
			to[i]=1;
		}
		for(int i=m;i>=1;--i){
			if(B[x[i]]==B[y[i]]){
				seq[i]=0;
				continue;
			}
			if(to[B[x[i]]]>=2){
				seq[i]=1;
				to[B[x[i]]]--;
				B[x[i]]=B[y[i]];
				to[B[y[i]]]++; 
			}else{
				seq[i]=0;
				to[B[y[i]]]--;
				B[y[i]]=B[x[i]];
				to[B[x[i]]]++;
			}
		}
		for(int i=1;i<=m;++i)
		    if(seq[i]==0)printf("^");
		    else printf("v");
    }
    return 0;
}