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:
// Written by newbiechd
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <vector>

const int BUFFER_SIZE = 1 << 25 | 1;
struct InputOutputStream {
  char ibuf[BUFFER_SIZE], obuf[BUFFER_SIZE], *s, *oh;
  InputOutputStream() : s(ibuf), oh(obuf) {
    ibuf[fread(ibuf, 1, BUFFER_SIZE, stdin)] = '\0';
  }
  ~InputOutputStream() { fwrite(obuf, 1, oh - obuf, stdout); }
  template <typename T>
  inline InputOutputStream &operator>>(T &x) {
    while (!isdigit(*s)) ++s;
    for (x = 0; isdigit(*s); ++s)
      x = x * 10 + (*s ^ '0');
    return *this;
  }
}IO;

template <typename lhs, typename rhs>
inline lhs min(const lhs &x, const rhs &y) { return x < y ? x : y; }
template <typename lhs, typename rhs>
inline lhs max(const lhs &x, const rhs &y) { return x > y ? x : y; }

const int maxN = 50003, maxM = 100003;
int u[maxM], v[maxM], cnt[maxN], tar[maxN];
std::bitset<maxN> a[maxN];
char ans[maxM];

int main() {
  int n, m, T;
  IO >> n >> m >> T;
  if (T == 1) {
    for (int i = 1; i <= n; ++i)
      a[i].set(i);
    for (int i = 1; i <= m; ++i)
      IO >> u[i] >> v[i], a[u[i]] = a[v[i]] = a[u[i]] | a[v[i]];
    for (int i = 1; i <= n; ++i)
      if (int(a[i].count()) == n) {
        static std::bitset<maxN> b;
        b.set(i);
        for (int j = m; j; --j)
          if (b[u[j]])
            b.set(v[j]), ans[j] = '^';
          else
            b[u[j]] = b[u[j]] | b[v[j]], ans[j] = 'v';
        printf("%s\n", ans + 1);
        return 0;
      }
    printf("-1\n");
  }
  else {
    if (n < 3) {
      printf("-1\n");
      return 0;
    }
    for (int i = 1; i <= m; ++i)
      IO >> u[i] >> v[i];
    for (int i = 1; i <= n; ++i)
      ++cnt[i], tar[i] = i;
    for (int i = m; i; --i) {
      if (cnt[tar[u[i]]] > 1)
        ans[i] = 'v', ++cnt[tar[v[i]]], --cnt[tar[u[i]]], tar[u[i]] = tar[v[i]];
      else
        ans[i] = '^', ++cnt[tar[u[i]]], --cnt[tar[v[i]]], tar[v[i]] = tar[u[i]];
    }
    printf("%s\n", ans + 1);
  }
  return 0;
}