Problem:
Score : 1600 points

Problem Statement

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).

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.

Partial Score

800 points will be awarded for passing the testset satisfying T = 1.

800 points will be awarded for passing the testset satisfying T = 2.

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.

Sample Input 1

4 5 1
1 3
2 4
1 2
3 4
2 3

Sample Output 1

^^^^^

This state is uniforming: regardless of the starting wire, the token ends up at wire 1.

Sample Input 2

4 5 2
1 3
2 4
1 2
3 4
2 3

Sample Output 2

v^^^^

This state is non-uniforming: depending on the starting wire, the token might end up at wire 1 or wire 2.

Sample Input 3

3 1 1
1 2

Sample Output 3

-1

A uniforming state doesn't exist.

Sample Input 4

2 1 2
1 2

Sample Output 4

-1

A non-uniforming state doesn't exist.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
//#define MOD 998244353
#define NUM 200010

void swap(int* a,int* b){int tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}
LL GCD(LL a,LL b){LL c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int POW(int a,int n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return (int)ans;}
int icmp(const void* a,const void* b){return *(int*)a-*(int*)b;}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int scmp(const void* a,const void* b){return strcmp(*(char**)a,*(char**)b);}
LL fac[NUM],finv[NUM],inv[NUM];
void comi(){int i;fac[0]=fac[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fac[i]=fac[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
int com(int n,int k){if(n<k||n<0||k<0)return 0;return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;}

int main(){
  int N,M,T,i,a=0;
  scanf("%d%d%d",&N,&M,&T);
  int x[M],y[M],flag[N+1];
  FOR(i,1,N+1)flag[i]=0;
  char ans[M];
  FOR(i,0,M)scanf("%d%d",&x[i],&y[i]);
  flag[x[M-1]]=1,flag[y[M-1]]=1;
  ans[M-1]='^';
  for(i=M-2;i>=0;i--){
    if(flag[x[i]]==1){
      flag[y[i]]=1;
      ans[i]='^';
    }
    else if(flag[y[i]]==1){
      flag[x[i]]=1;
      ans[i]='v';
    }
    else ans[i]='v';
  }
  FOR(i,1,N+1)if(flag[i]==0)a=1;

  if(a==0)FOR(i,0,M)printf("%c",ans[i]);
  else printf("-1");
	return 0;
}