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:
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int N=5e4+5,M=1e5+5;
int a[M],b[M],n,m,T;
bool col[M];

namespace sub1{

	bool ok[N];
	bitset<N> can[N];
	
	void main(){
		for(int i=1;i<=n;i++) can[i][i]=1;
		for(int i=1;i<=m;i++){
			can[a[i]]=can[b[i]]=can[a[i]]|can[b[i]];
			if(can[a[i]].count()==n){
				ok[a[i]]=ok[b[i]]=1;
				for(int j=i-1;j;j--) if(ok[a[j]]||ok[b[j]]){
					if(ok[a[j]]) col[j]=1;
					ok[a[j]]=ok[b[j]]=1;
				}
				for(int i=1;i<=m;i++) putchar(col[i]?'^':'v');
				return ;
			}
		}
		write(-1);
	}
}

namespace sub2{
	
	int sz[N],to[N];
	bool flag;
	
	void main(){
		for(int i=1;i<=n;i++) to[i]=i,sz[i]=1;
		for(int i=m;i;i--){
			if(max(sz[to[a[i]]]-1,sz[to[b[i]]]+1)<max(sz[to[a[i]]]+1,sz[to[b[i]]]-1)) sz[to[a[i]]]--,sz[to[a[i]]=to[b[i]]]++;
			else col[i]=1,sz[to[b[i]]]--,sz[to[b[i]]=to[a[i]]]++;;
		}
		for(int i=1;i<=n;i++) flag|=sz[i]==n;
		if(flag) write(-1);
		else for(int i=1;i<=m;i++) putchar(col[i]?'^':'v');
	}
}

signed main(){
	read(n,m,T);
	for(int i=1;i<=m;i++) read(a[i],b[i]);
	if(T==1) sub1::main();
	else sub2::main();
}