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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se seoncd
#define pb push_back
#define mp make_pair
using namespace std;
typedef bitset<50001> bits;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=100005;
int x[N],y[N],v[N],ans[N],n,m,T,a,b;
bits f[N>>1];
vi id[N];

int main(){
	read(n),read(m),read(T);
	if(T==1){
		rep(i,1,n) f[i].set(i);
		rep(i,1,m){
			read(x[i]),read(y[i]);
			f[x[i]]|=f[y[i]],f[y[i]]|=f[x[i]];
		}
		rep(k,1,n){
			if(f[k].count()<n) continue;
			v[k]=1;
			per(i,m,1){
				if(v[x[i]]) v[y[i]]=1,ans[i]=1;
				else v[x[i]]=v[y[i]];
			}
			rep(i,1,m) putchar(ans[i]?'^':'v');
			return 0;
		}
		puts("-1");
		return 0;
	}
	if(n==2){
		puts("-1");
		return 0;
	}
	rep(i,1,m){
		read(x[i]),read(y[i]);
		id[x[i]].pb(i),id[y[i]].pb(i);
	}
	rep(i,1,n){
		reverse(id[i].begin(),id[i].end());
	}
	a=x[1],b=(y[1]==n?n-1:y[1]+1);
	if(a==b) a--;
	rep(i,1,m){
		int ca=(a==x[i]||a==y[i]);
		int cb=(b==x[i]||b==y[i]);
		if(!ca&&!cb) continue;
		assert(ca+cb==1);
		while(!id[a].empty()&&id[a].back()<=i) id[a].pop_back();
		while(!id[b].empty()&&id[b].back()<=i) id[b].pop_back();
		int ka=id[a].empty()?-1:id[a].back();
		int kb=id[b].empty()?-1:id[b].back();
		int k=min(ka,kb),f=(k==-1||a!=x[k]||b!=y[k]);
		if(cb) swap(a,b);
		if((y[i]==a)^f) a=x[i],ans[i]=1;
		else a=y[i];
		if(a>b) swap(a,b);
	}
	rep(i,1,m) putchar(ans[i]?'^':'v');
	return 0;
}