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;
#define ll long long
char buf[1<<20],*p1,*p2;
#define GC (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
template<class T> inline void read(T &n){
	char ch=GC;T w=1,x=0;
	while(!isdigit(ch)){if(ch=='-') w=-1;ch=GC;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=GC;}
	n=x*w;
}
const int maxn=50005;
const int maxm=100005;
int T,n,m,x[maxm],y[maxm],res[maxm];
namespace solve1{
	bitset <maxn> f[maxn];
	int ok[maxn];
	void main(){
		bitset <maxn> tmp;
		for(int i=1;i<=n;i++) f[i][i]=1;
		for(int i=m;i>=1;i--){
			tmp=f[x[i]]|f[y[i]];
			f[x[i]]=f[y[i]]=tmp;
		}
		int pos=-1;
		tmp=f[1];
		for(int i=2;i<=n;i++) tmp&=f[i];
		for(int i=1;i<=n;i++) 
			if(tmp[i]){
				pos=i;
				break;
			}
		if(pos==-1) puts("-1"),exit(0);
		ok[pos]=1;
		for(int i=m;i>=1;i--){
			if(ok[x[i]]^ok[y[i]]) res[i]=ok[x[i]];
			ok[x[i]]=ok[y[i]]=(ok[x[i]]|ok[y[i]]);
		}
		for(int i=1;i<=m;i++) res[i]?putchar('^'):putchar('v');
	}
}
namespace solve2{
	int ed[maxn],cnt[maxn];
	void main(){
		if(n==2) puts("-1"),exit(0);
		for(int i=1;i<=n;i++) ed[i]=i,cnt[i]=1;
		for(int i=m;i>=1;i--){
			if(ed[x[i]]==ed[y[i]]) continue;
			if(cnt[ed[x[i]]]>=2){
				cnt[ed[x[i]]]--;
				cnt[ed[y[i]]]++;
				ed[x[i]]=ed[y[i]];
			}
			else{
				cnt[ed[x[i]]]++;
				cnt[ed[y[i]]]--;
				ed[y[i]]=ed[x[i]];
				res[i]=1;
			}
		}
		for(int i=1;i<=m;i++) res[i]?putchar('^'):putchar('v');
	}
}
int main(){
	read(n),read(m),read(T);
	for(int i=1;i<=m;i++) read(x[i]),read(y[i]); 
	if(T==1) solve1::main();
	if(T==2) solve2::main();
	return 0;
}