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
#include <bits/stdc++.h>
 
#include <unistd.h>
 
using namespace std;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
 
 
typedef long long ll;
#define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define All(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
const ll MOD = 1000000007;
//const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'

void solve2(int n, int m) {
  if (n == 2) {
    cout << -1 << endl;
    return;
  }
  vi x(m), y(m);
  rep (i, m) {
    cin >> x[i] >> y[i];
    x[i]--; y[i]--;
  }

  vi num(n), w(n);
  rep (i, n) {
    num[i] = 1;
    w[i] = i;
  }

  string s;
  rrep (i, m) {
    if (num[w[x[i]]] >= 2) {
      num[w[x[i]]]--;
      num[w[y[i]]]++;
      s += 'v';
      w[x[i]] = w[y[i]];
    }
    else if (num[w[y[i]]] >= 2) {
      num[w[y[i]]]--;
      num[w[x[i]]]++;
      s += '^';
      w[y[i]] = w[x[i]];
    }
    else {
      num[w[x[i]]]--;
      num[w[y[i]]]++;
      s += 'v';
      w[x[i]] = w[y[i]];
    }
  }

  reverse(All(s));
  cout << s << endl;
}

void solve1(int n, int m) {
  vi x(m), y(m);
  rep (i, m) {
    cin >> x[i] >> y[i];
    x[i]--; y[i]--;
  }

  vector<bitset<50001>> now(n);
  rep (last, n) {
    now[last][last] = true;
  }
  //DEBUG_VEC(now);
  rrep (i, m) {
    now[x[i]] = now[y[i]] = now[x[i]] | now[y[i]];
    // rep (last, n) {
    //   now[x[i]][last] = now[y[i]][last] = now[x[i]][last] | now[y[i]][last];
    // }
    //DEBUG_VEC(now);
  }

  bitset<50001> ok;
  ok.set();
  rep (i, n) {
    ok &= now[i];
  }

  int w = -1;
  rep (i, n) {
    if (ok[i]) {
      w = i;
      break;
    }
  }
  
  if (w == -1) {
    cout << -1 << endl;
    return;
  }

  int last = w;

  bitset<50001> now2;
  now2[last] = true;
  string ans;
  rrep (i, m) {
    if (now2[x[i]] or now2[y[i]]) {
      if (not now2[x[i]]) {
        now2[x[i]] = true;
        ans += 'v';
      }
      else if (not now2[y[i]]) {
        now2[y[i]] = true;
        ans += '^';
      }
      else {
        ans += 'v';
      }
    }
    else {
      ans += 'v';
    }
  }
  reverse(All(ans));
  cout << ans << endl;
}

signed main() {
  fio();
  int n, m, t;
  cin >> n >> m >> t;
  if (t == 2) {
    solve2(n, m);
  }
  else {
    solve1(n, m);
  }
}
