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 range(i, m, n) for(int i = m; i < n; i++)
#define husk(i, m, n) for(int i = m; i > n; i--)

template <typename T>
void debug_out(T t) {
  cerr << t;
}

template <typename A, typename B>
void debug_out(pair<A, B> u) {
  cerr << "(" << u.first << " " << u.second << ")";
}

template <typename T>
void debug_out(vector<T> t) {
  int sz = t.size();
  for(int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if(i != sz - 1) cerr << ", ";
  }
}

template <typename T>
void debug_out(vector<vector<T>> t) {
  int sz = t.size();
  for(int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if(i != sz - 1) cerr << endl;
  }
}

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  debug_out(H);
  cerr << " ";
  debug_out(T...);
}

template <typename T>
void debug_out(set<T> a) {
  vector<T> _a;
  for(T t : a) _a.push_back(t);
  debug_out(_a);
}

#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:" << endl, debug_out(__VA_ARGS__), cerr << endl;

int n, m, t;

int main() {
  cin >> n >> m >> t;
  vector<int> x(m);
  vector<int> y(m);
  vector<vector<int>> g(n);
  range(i, 0, m) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    g[x[i]].push_back(i);
    g[y[i]].push_back(i);
  }
  if(t == 2) {
    if(n == 2) {
      cout << -1;
      return 0;
    }
    string ans(m, '^');
    vector<int> pos;
    range(i, 0, m) if(y[i] <= 2) pos.push_back(i);
    range(i, 0, (int) pos.size() - 1) {
      int sx = x[pos[i + 1]];
      int sy = y[pos[i + 1]];
      int cx = x[pos[i]];
      int cy = y[pos[i]];
      if(cx != sx && cx != sy) {
        ans[pos[i]] = '^';
      }
      if(cy != sx && cy != sy) {
        ans[pos[i]] = 'v';
      }
    }
    cout << ans;
  } else {
    const int N = 50010;
    vector<bitset<N>> pos(n);
    range(i, 0, n) pos[i][i] = 1;
    range(i, 0, m) {
      pos[x[i]] |= pos[y[i]];
      pos[y[i]] = pos[x[i]];
    }
    int root = -1;
    range(i, 0, n) {
      if((int) pos[i].count() == n) root = i;
    }
    if(root == -1) {
      cout << -1;
      return 0;
    }

    int cnt = 0;
    vector<int> l, r;
    vector<int> wire;
    vector<int> foo(n);
    range(i, 0, n) {
      foo[i] = cnt;
      l.push_back(-1);
      r.push_back(g[i][0]);
      cnt++;
      wire.push_back(i);

      range(j, 0, (int) g[i].size() - 1) {
        l.push_back(g[i][j]);
        r.push_back(g[i][j + 1]);
        cnt++;
        wire.push_back(i);
      }

      l.push_back(g[i].back());
      r.push_back(m);
      cnt++;
      wire.push_back(i);
    }

    vector<int> nxt(cnt, -1);
    vector<int> par(cnt, -1);
    vector<vector<int>> rev(cnt);
    range(i, 0, cnt) {
      if(r[i] != m) {
        nxt[i] = i + 1;
        rev[nxt[i]].push_back(i);
        int id = x[r[i]] + y[r[i]] - wire[i];
        par[i] = foo[id] + (lower_bound(g[id].begin(), g[id].end(), r[i]) - g[id].begin()) + 1;
        assert(l[par[i]] == r[i]);
        rev[par[i]].push_back(i);
      }
    }

    vector<int> was(cnt);
    function<void(int)> dfs = [&](int v) {
      was[v] = 1;
      for(int i : rev[v]) {
        if(was[i]) continue;
        dfs(i);
      }
    };
    dfs(foo[root] + g[root].size());

    string ans(m, '^');
    range(i, 0, cnt) {
      if(!was[i]) continue;
      if(r[i] == m) continue;
      assert(was[par[i]] || was[nxt[i]]);
      int id = r[i];
      if(was[par[i]]) ans[id] = (wire[par[i]] > wire[i] ? 'v' : '^');
      if(was[nxt[i]]) ans[id] = (wire[par[i]] > wire[i] ? '^' : 'v');
    }
    cout << ans;
  }
  return 0;
}