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 <stdio.h>
#include <algorithm>
#include <assert.h>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <deque>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <time.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <chrono>
#include <random>
#include <time.h>
#include <fstream>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;++i)
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep3(i,a,b) for(ll i=a;i>=b;i--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pq priority_queue<int>
#define pqg priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define eb emplace_back
#define ep emplace
#define vec vector<int>
#define vecll vector<ll>
#define vecpii vector<pii>
#define endl "\n"
#define all(c) begin(c),end(c)
using namespace std;
int in() {int x;scanf("%d",&x);return x;}
ll lin() {ll x;scanf("%lld",&x);return x;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
void print(vec v){for(auto e:v)cout<<e<<" ";cout<<endl;}
void print(vecll v){for(auto e:v)cout<<e<<" ";cout<<endl;}
void print(vector<vec> v){for(auto e:v){for(auto ee:e)cout<<ee<<" ";cout<<endl;}}
void print(map<int,int> mp){for(auto e:mp)cout<<e.first<<" "<<e.second<<endl;cout<<endl;}
const int INF=1e6;
const ll int MOD=998244353;
#define N 50005
bitset<N> a[N],p;
int main(){
    int n=in(),m=in(),t=in();
    if(t==1){
        rep(i,n+1)a[i].reset();
        rep(i,n)a[i][i]=1;
        vec x,y;
        rep(i,m)x.pb(in()-1),y.pb(in()-1);
        reverse(all(x)),reverse(all(y));
        rep(i,m){
            a[x[i]]|=a[y[i]];
            a[y[i]]|=a[x[i]];
        }
        rep(i,n)a[0]&=a[i];
        int s=-1;
        rep(i,n){
            if(a[0][i])s=i;
        }
        if(s==-1){
            cout<<-1;return 0;
        }
        p.reset();
        p[s]=1;
        string ans;
        rep(i,m){
            if(p[x[i]]&&!p[y[i]])ans.pb('^');
            else ans.pb('v');
            if(p[x[i]])p[y[i]]=1;
            else if(p[y[i]])p[x[i]]=1;
        }
        reverse(all(ans));
        cout<<ans<<endl;
    }
    else{
        if(n==2) cout<<-1;
        else{
            string ans;
            vec x,y;
            rep(i,m){
                x.pb(in());
                y.pb(in());
            }
            reverse(all(x));
            reverse(all(y));
            vec mem(4);
            rep(i,m){
                if(y[i]>3)ans.pb('^');
                else{
                    if(mem[x[i]])ans.pb('v');
                    else ans.pb('^');
                    mem=vec(4);
                    mem[x[i]]=mem[y[i]]=1;
                }
            }
            reverse(all(ans));
            cout<<ans;
        }
    }
}