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:
#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define DB double
#define BS bitset
#define VE vector
#define VL vector<LL>
#define VP vector<pair<LL,LL>>
#define VVL vector<vector<LL>>
#define PQ priority_queue<LL>
#define PQS priority_queue<LL, vector<LL>, greater<LL>>
#define PB push_back
#define POB pop_back
#define PF push_front
#define POF pop_front
#define MP make_pair
#define TS to_string
#define TU to_ullong
#define FOR(i,a,n) for(i=a;i<n;i++)
#define rep(i,n) FOR(i,0,n)
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define SORT(a) sort(ALL(a))
#define REV(a) reverse(ALL(a))
#define UB(a,n) *upper_bound(ALL(a),n)
#define LB(a,n) *lower_bound(ALL(a),n)
#define INF 1145141919810364364
#define PI 3.14159265358979
#define MOD 1000000007
//#define MOD 998244353
#define ERR 0.00000000000001
#define NUM 200010

void yn(LL a){if(a)printf("Yes\n");else printf("No\n");}
void YN(LL a){if(a)printf("YES\n");else printf("NO\n");}
LL pwmn(LL a,LL n){LL ans=1;while(ans<a)ans*=n;return ans;}
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);}
LL mod(LL a,LL m){if(a<0)return a%m+m;else return a%m;}
LL DIV(LL a,LL d){LL m=MOD,x=1,y=0,k;while(m){k=d/m;d-=k*m;swap(m,d);x-=k*y;swap(x,y);}return mod(a*mod(x,MOD),MOD);}
LL FAC(LL a){LL i,ans=1;FOR(i,1,a+1){ans*=i;if(MOD>0&&ans>MOD)ans%=MOD;}return ans;}
LL POW(LL a,LL n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return ans;}
LL fact[NUM],finv[NUM],inv[NUM];
void comi(){LL i;fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}
template<typename T>bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<typename T>bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

int main(){
  LL N,M,T,i,j,a=0;
  VL x(100005),y(100005),kf(50005),kn(50005);
  BS<100005> ans;
  VE<BS<50005>> f(50005);
  cin>>N>>M>>T;
  rep(i,M)cin>>x[i]>>y[i];
  if(T==1){
    FOR(i,1,N+1)f[i][i]=1;
    rep(i,M)f[x[i]]=f[y[i]]=f[x[i]]|f[y[i]];
    FOR(i,1,N+1){
      if(f[i].count()!=N)continue;
      f[i].reset();
      f[i][i]=1;
      for(j=M-1;j>=0;j--){
        if(f[i].test(x[j]))f[i][y[j]]=1,ans[j]=1;
        else if(f[i].test(y[j]))f[i][x[j]]=1,ans[j]=0;
        else ans[j]=0;
      }
      if(f[i].count()==N)a=1;
      if(a==1){
        rep(i,M)cout<<(ans.test(i)?'^':'v');
        return 0;
      }
    }
    cout<<-1<<endl;
  }else{
    FOR(i,1,N+1)kf[i]=i,kn[i]=1;
    for(i=M-1;i>=0;i--){
      if(kn[kf[x[i]]]>kn[kf[y[i]]]){
        ans[i]=0;
        kn[kf[x[i]]]--,kn[kf[y[i]]]++;
        kf[x[i]]=kf[y[i]];
      }else{
        ans[i]=1;
        kn[kf[x[i]]]++,kn[kf[y[i]]]--;
        kf[y[i]]=kf[x[i]];
      }
    }
    FOR(i,1,N+1)if(kn[i]>0)a++;
    if(a>1)rep(i,M)cout<<(ans.test(i)?'^':'v');
    else printf("-1");
  }

	return 0;
}