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;
void E(){puts("-1");exit(0);}
const int N=1e5+5;
int T,n,m,i,j,a[N],x[N],y[N];
char c[N];
namespace T1{
bitset<50005>b[N],all;
int lst[N];
bool bb[N];
inline void main(){
	for(i=1;i<=n;++i)all.set(i);
	for(i=1;i<=m;++i){
		if(!lst[x[i]])b[i].set(x[i]);else b[i]|=b[lst[x[i]]];
		if(!lst[y[i]])b[i].set(y[i]);else b[i]|=b[lst[y[i]]];
		lst[x[i]]=lst[y[i]]=i;
		if(b[i]==all)break;
	}
	if(i>m)E();
	for(j=m;j>i;--j)c[j]='^';
	bb[x[i]]=bb[y[i]]=1;
	for(j=i;j;--j)if(bb[x[j]])c[j]='^',bb[y[j]]=1;
		else c[j]='v',bb[x[j]]|=bb[y[j]];
}
}
namespace T2{
int u[N],v[N],z;
bool fl[N];
void link(int x,int y){a[x]=y;a[y]=x;}
void setv(int i,int z){c[i]=(x[i]<y[i])^(z==y[i])?'^':'v';}
inline void main(){
	if(n==2 && m>=1)E();
	memset(a+1,0,n<<2);
	for(i=1;i<=m;++i){
		if(!a[x[i]] && !a[y[i]]){
			u[i]=-1;link(x[i],y[i]);
		}else if(a[x[i]] && a[y[i]]){
			if(a[x[i]]!=y[i]){
				u[i]=a[x[i]];v[i]=a[y[i]];
				link(u[i],v[i]);link(x[i],y[i]);
			}else u[i]=0;
		}else{
			if(a[y[i]])swap(x[i],y[i]);
			u[i]=-2;v[i]=x[i];a[a[x[i]]]=0;link(x[i],y[i]);
		}
	}
	memset(fl+1,1,n);memset(c+1,0,m+2);
	for(i=1;i<=n;++i)if(a[i]>i)fl[i]=0;
	for(i=m;i;--i){
		if(u[i]>0){
			if(fl[u[i]] && fl[x[i]])fl[x[i]]=0,fl[y[i]]=1,setv(i,x[i]);
				else if(fl[u[i]] && fl[y[i]])setv(i,y[i]);
					else if(fl[v[i]] && fl[x[i]])setv(i,x[i]);
						else fl[x[i]]=1,fl[y[i]]=0,setv(i,y[i]);
		}
		if(u[i]==0 || u[i]==-1)setv(i,fl[x[i]]?x[i]:y[i]);
		if(u[i]==-2){
			setv(i,fl[x[i]]?x[i]:y[i]);
			fl[v[i]]=0;fl[x[i]^y[i]^v[i]]=1;
		}
	}
}
}
int main(){
	scanf("%d%d%d",&n,&m,&T);for(i=1;i<=m;++i)scanf("%d%d",x+i,y+i);
	if(T==2)T2::main();else T1::main();
	puts(c+1);
}