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>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define IT iterator
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
#define debug puts("wzpakking")
#define y1 ysghysgsygsh
using namespace std;

const int N=50005;
int n,m,T,x[N*2],y[N*2];
namespace solver1{
	uint B[N][N/32+5],fu[N/32+5];
	char ans[N*2];
	int vis[N];
	void solve(){
		For(i,1,m) x[i]--,y[i]--;
		For(i,0,n-1) B[i][i>>5]|=(1u<<(i&31));
		For(i,0,n-1) fu[i>>5]|=(1u<<(i&31));
		For(i,1,m) For(j,0,n/32)
			B[x[i]][j]=B[y[i]][j]=B[x[i]][j]|B[y[i]][j];
		int sp=-1;
		For(i,0,n-1){
			bool fl=0;
			For(j,0,n/32)
				if (fu[j]!=B[i][j])
					fl=1;
			if (!fl) sp=i;
		}
		if (sp==-1) return puts("-1"),void(0);
		vis[sp]=1;
		Rep(i,m,1){
			if (vis[x[i]]==vis[y[i]]) ans[i]='^';
			else if (vis[x[i]]) ans[i]='^';
			else ans[i]='v';
			vis[x[i]]=vis[y[i]]=vis[x[i]]|vis[y[i]];
		}
		printf("%s\n",ans+1);
	}
}
namespace solver2{
	char ans[N*2];
	int be[N*2],cnt[N*2];
	void solve(){
		if (n==2) return puts("-1"),void(0);
		For(i,1,n) be[i]=i,cnt[i]=1;
		Rep(i,m,1){
			int X=be[x[i]],Y=be[y[i]];
			if (cnt[X]<cnt[Y]){
				ans[i]='^';
				cnt[X]++;
				cnt[Y]--;
				be[y[i]]=be[x[i]];
			}
			else{
				ans[i]='v';
				cnt[X]--;
				cnt[Y]++;
				be[x[i]]=be[y[i]];
			}
		}
		printf("%s\n",ans+1);
	}
}
int main(){
	scanf("%d%d%d",&n,&m,&T);
	For(i,1,m) scanf("%d%d",&x[i],&y[i]);
	if (T==1) solver1::solve();
	else solver2::solve(); 
}
