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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<bitset>
#include<set>
#define N (100010)
#define P ()
#define M ()
#define inf (0x7f7f7f7f)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
namespace fastIO1{
	inline char read(){
		static const int IN_LEN=1000000;
		static char buf[IN_LEN],*s,*t;
		return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
	}
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=read();!isdigit(c);c=read()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
	inline char readc(char &c){
		for(c=read();!isalpha(c)&&!isdigit(c);c=read())
		if(c==-1)return 0;
	}
	const int OUT_LEN = 10000000;
	char obuf[OUT_LEN],*ooh=obuf;
	inline void print(char c){
		if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
		*ooh++=c;
	}
	template<class T>
	inline void print(T x){
		static int buf[30],cnt;
		if(x==0)print('0');
		else{
			if(x<0)print('-'),x=-x;
			for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
			while(cnt)print((char)buf[cnt--]);
		}
	}
	inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
}
namespace fastIO2{
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=getchar();!isdigit(c);c=getchar()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=getchar())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
}
using namespace fastIO1;
int n,m,tp,a[N],b[N],cnt[N],to[N]; bool yes[N],d[N];
bitset<50010> bs[50010],qwq;
int main(){
	read(n),read(m),read(tp); qwq.set();
	for(int i=1;i<=m;i++)read(a[i]),read(b[i]);
	if(tp==1){
		int ans=-1;
		for(int i=1;i<=n;i++)bs[i][i]=1;
		for(int i=m;i;i--)bs[a[i]]=bs[b[i]]=bs[a[i]]|bs[b[i]];
		for(int i=1;i<=n;i++)qwq&=bs[i];
		for(int i=1;i<=n;i++)if(qwq[i]){ans=i;break;}
		if(ans==-1){cout<<ans<<endl;return 0;}
		yes[ans]=1;
		for(int i=m;i;i--){
			if(yes[b[i]])d[i]=1;
			yes[a[i]]=yes[b[i]]=yes[a[i]]|yes[b[i]];
		}
		for(int i=1;i<=m;i++)
		putchar(d[i]?'v':'^');
	}
	else{
		if(n==2){puts("-1");return 0;}
		for(int i=1;i<=n;i++)cnt[i]=1,to[i]=i;
		for(int i=m;i;i--){
			if(cnt[to[a[i]]]>1)d[i]=1,cnt[to[a[i]]]--,cnt[to[b[i]]]++,to[a[i]]=to[b[i]];
			else{
				cnt[to[a[i]]]++,cnt[to[b[i]]]--,to[b[i]]=to[a[i]];
			}
		}
		for(int i=1;i<=m;i++)
		putchar(d[i]?'v':'^');
	}
}