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 <bits/stdc++.h>

using namespace std;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n, m, t, x[100100], y[100100];
bitset<50000> r[50000];

void reconstruct(int i) {
  bitset<50000> c;
  c[i] = true;
  string res;
  for (int j = m - 1; j >= 0; j--) {
    if (c[x[j]])
      res.push_back('^'), c[y[j]] = true;
    else if (c[y[j]])
      res.push_back('v'), c[x[j]] = true;
    else
      res.push_back('^');
  }
  reverse(res.begin(), res.end());
  cout << res << "\n";
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n >> m >> t;
  for (int i = 0; i < m; i++)
    cin >> x[i] >> y[i], x[i]--, y[i]--;
  if (t == 1) {
    for (int i = 0; i < n; i++)
      r[i][i] = true;
    for (int i = m - 1; i >= 0; i--)
      r[x[i]] |= r[y[i]], r[y[i]] = r[x[i]];
    bitset<50000> res;
    res.set();
    for (int i = 0; i < n; i++)
      res &= r[i];
    if (res.any()) {
      int i = 0;
      while (!res[i])
        i++;
      reconstruct(i);
    } else
      cout << -1 << "\n";
  } else {
    if (n == 2) {
      cout << -1 << "\n";
      return 0;
    }
    int a = 0, b = 1, c = 2, l = -1;
    string res;
    for (int i = 0; i < m; i++) {
      if (!(x[i] == a || x[i] == b || x[i] == c))
        res.push_back('v');
      else if (!(y[i] == a || y[i] == b || y[i] == c))
        res.push_back('^');
      else if (l == -1) {
        if (b == x[i])
          swap(a, b);
        else if (c == x[i])
          swap(a, c);
        if (c == y[i])
          swap(b, c);
        l = i;
        res.push_back('^');
      } else if (x[i] == c || y[i] == c) {
        if (x[i] == a || y[i] == a) {
          res[l] = 'v';
          swap(a, b);
        }
        swap(a, c);
        if (a > b)
          swap(a, b);
        res.push_back('^');
        l = i;
      } else {
        res.push_back('^');
        l = i;
      }
    }
    cout << res << "\n";
  }
}