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;

struct data {
  long long x, y;
};

long long n, m, t;
vector<data> v;
set<long long> st;
vector<bitset<50000>> bv;

string solve();

int main() {
  cin >> n >> m >> t;
  v.resize(m);
  for(int i = 0; i < m; ++i) {
    cin >> v[i].x >> v[i].y;
    --v[i].x, --v[i].y;
  }
  cout << solve() << endl;
  return 0;
}

string solve() {
  if(t == 1) {
    bitset<50000> bs;
    for(int i = n; i < 50000; ++i) bs[i] = 1;
    bv.assign(n, bs);
    for(int i = 0; i < n; ++i) bv[i][i] = 1;
    for(int i = 0; i < m; ++i) {
      bv[v[i].x] |= bv[v[i].y];
      bv[v[i].y] = bv[v[i].x];
    }
    int id = -1;
    for(int i = 0; i < n; ++i)
      if(bv[i].all()) id = i;
    if(id == -1) return "-1";
    string res;
    bs[id] = 1;
    reverse(v.begin(), v.end());
    for(int i = 0; i < m; ++i) {
      if(bs[v[i].x] && !bs[v[i].y])
        res += '^';
      else
        res += 'v';
      bs[v[i].x] = bs[v[i].x] | bs[v[i].y];
      bs[v[i].y] = bs[v[i].x];
    }
    reverse(res.begin(), res.end());
    return res;
  }
  if(n == 2) return "-1";
  vector<long long> bs;
  map<long long, long long> mp;
  bs.resize(n);
  for(int i = 0; i < n; ++i) bs[i] = i, mp[i] = 1;
  reverse(v.begin(), v.end());
  string res;
  for(int i = 0; i < m; ++i) {
    if(mp[bs[v[i].x]] > 1) {
      --mp[bs[v[i].x]];
      ++mp[bs[v[i].y]];
      bs[v[i].x] = bs[v[i].y];
      res += 'v';
    }
    else {
      --mp[bs[v[i].y]];
      ++mp[bs[v[i].x]];
      bs[v[i].y] = bs[v[i].x];
      res += '^';
    }
  }
  reverse(res.begin(), res.end());
  return res;
}