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 _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
template <typename T> using posteriority_queue = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
// const int dy[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
int popcount(int val) { return __builtin_popcount(val); }
int popcountll(ll val) { return __builtin_popcountll(val); }
template <typename T> void unique(vector<T> &a) { a.erase(unique(ALL(a)), a.end()); }
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;

int main() {
  int n, m, t; cin >> n >> m >> t;
  vector<int> x(m), y(m);
  REP(i, m) {
    cin >> x[i] >> y[i]; --x[i]; --y[i];
  }
  if (t == 1) {
    vector<bitset<50000> > bit(n, bitset<50000>(0));
    REP(i, n) bit[i].set(i);
    REP(i, m) bit[x[i]] = bit[y[i]] = bit[x[i]] | bit[y[i]];
    REP(i, n) {
      if (bit[i].count() < n) continue;
      vector<char> ans(m);
      vector<bool> visited(n, false);
      visited[i] = true;
      for (int j = m - 1; j >= 0; --j) {
        if (visited[x[j]]) {
          ans[j] = '^';
          visited[y[j]] = true;
        } else if (visited[y[j]]) {
          ans[j] = 'v';
          visited[x[j]] = true;
        } else {
          ans[j] = '^';
        }
      }
      REP(j, m) cout << ans[j];
      cout << '\n';
      return 0;
    }
    cout << "-1\n";
    return 0;
  } else if (t == 2) {
    if (n == 2) {
      cout << "-1\n";
      return 0;
    }
    vector<int> b(n), cnt(n, 1);
    iota(ALL(b), 0);
    vector<char> ans(m);
    for (int i = m - 1; i >= 0; --i) {
      if (cnt[b[x[i]]] >= 2) {
        ans[i] = 'v';
        --cnt[b[x[i]]];
        b[x[i]] = b[y[i]];
        ++cnt[b[x[i]]];
      } else {
        ans[i] = '^';
        --cnt[b[y[i]]];
        b[y[i]] = b[x[i]];
        ++cnt[b[y[i]]];
      }
    }
    REP(i, m) cout << ans[i];
    cout << '\n';
    return 0;
  }
}