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:
/*
Author: QAQAutomaton
Lang: C++
Code: E.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		memset(&inf,0x3f,sizeof(inf));
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
namespace solve0{
	bitset<50005> f[50005];
	bitset<100005>ans;
	int fa[50005],ok[50005];
	int u[100005],v[100005];
	int find(int x){
		return x==fa[x]?x:fa[x]=find(fa[x]);
	}
	void main(int n,int m){
		for(int i=1;i<=n;++i)f[i][i]=1;
		for(int i=1;i<=m;++i){
			read(u[i],v[i]);
			f[u[i]]|=f[v[i]];
			f[v[i]]=f[u[i]];
		}
		for(int i=1;i<=n;++i){
			if(f[i].count()==n){
				for(int j=1;j<=n;++j)fa[j]=j;
				ok[i]=1;
				for(int j=m;j;--j){
					if(ok[find(u[j])]){
						ans[j]=0;
						fa[find(v[j])]=find(u[j]);
					}
					else if(ok[find(v[j])]){
						ans[j]=1;
						fa[find(u[j])]=find(v[j]);
					}
					else ans[j]=1; 
				}
				for(int j=1;j<=m;++j)if(ans[j])putc('v');else putc('^');
				putc('\n');
				return;
			}
		}
		write("-1\n");

	}
}
namespace solve{
	int u[100005],v[100005];
	int a[50005],cnt[50005];
	bitset<100005>ans;
	void main(int n,int m,int k){
		for(int i=1;i<=n;++i){
			a[i]=i;
			cnt[i]=1;
		}
		for(int i=1;i<=m;++i)read(u[i],v[i]);
		int w=n;
		for(int i=m;i;--i){
			if(cnt[a[u[i]]]>1){
				--cnt[a[u[i]]];++cnt[a[v[i]]];
				a[u[i]]=a[v[i]];
				ans[i]=1;
			}
			else if(cnt[a[v[i]]]>1){
				++cnt[a[u[i]]];--cnt[a[v[i]]];
				a[v[i]]=a[u[i]];
			}
			else{
				--cnt[a[u[i]]];++cnt[a[v[i]]];
				a[u[i]]=a[v[i]];--w;
			}
		}
		if(w>=k){
			for(int j=1;j<=m;++j)if(ans[j])putc('v');else putc('^');
			putc('\n');
		}
		else write("-1\n");
	}
}
signed main(){
#ifdef QAQAutoMaton 
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int n,m,t;
	read(n,m,t);
	if(t==1)solve0::main(n,m);
	else solve::main(n,m,2);

	return 0;
}