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:
/**
 *    author:  tourist
 *    created: 14.05.2019 17:29:26       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, t;
  cin >> n >> m >> t;
  vector<int> u(m), v(m);
  for (int i = 0; i < m; i++) {
    cin >> u[i] >> v[i];
    --u[i]; --v[i];
  }
  if (t == 1) {
    const int N = 50010;
    vector<bitset<N>> reach(n);
    for (int i = 0; i < n; i++) {
      reach[i][i] = 1;
    }
    for (int i = m - 1; i >= 0; i--) {
      reach[u[i]] |= reach[v[i]];
      reach[v[i]] = reach[u[i]];
    }
    bitset<N> all = reach[0];
    for (int i = 1; i < n; i++) {
      all &= reach[i];
    }
    cerr << "n = " << n << ", m = " << m << ", count = " << all.count() << '\n';
    for (int i = 0; i < n; i++) {
      if (all[i]) {
        cerr << "i = " << i << '\n';
        vector<bool> got(n, false);
        got[i] = true;
        string res(m, '^');
        for (int j = m - 1; j >= 0; j--) {
          if (got[u[j]] && !got[v[j]]) {
            res[j] = '^';
            got[v[j]] = true;
          }
          if (!got[u[j]] && got[v[j]]) {
            res[j] = 'v';
            got[u[j]] = true;
          }
        }
        cout << res << '\n';
        return 0;
      }
    }
    cout << -1 << '\n';
  } else {
    if (n == 2) {
      cout << -1 << '\n';
      return 0;
    }
    vector<int> ids;
    for (int i = 0; i < m; i++) {
      if (v[i] < 3) {
        ids.push_back(i);
      }
    }
    string res(m, '^');
    for (int it = (int) ids.size() - 2; it >= 0; it--) {
      int me = ids[it];
      int nxt = ids[it + 1];
      if (u[me] == u[nxt] && v[me] == v[nxt]) {
        res[me] = res[nxt];
      } else {
        if (u[me] == u[nxt] || u[me] == v[nxt]) {
          res[me] = 'v';
        }
      }
    }
    cout << res << '\n';
  }
  return 0;
}