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:
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

using ll = long long;
using ld = long double;
using ull = unsigned long long;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const int M = 120000;
const int N = 50010;

int n, m, tt;
int x[M];
int y[M];

bitset<N> bb[N];

int en[N];
char ans[M];

void solve1() {
	for (int i = 0; i < n; ++i) {
		bb[i][i] = 1;
	}
	for (int i = 0; i < m; ++i) {
		bb[x[i]] |= bb[y[i]];
		bb[y[i]] |= bb[x[i]];
	}

	for (int i = 0; i < n; ++i) {
		if (bb[i].count() == n) {
			en[i] = 1;
			for (int i = m - 1; i >= 0; --i) {
				if (en[x[i]]) {
					ans[i] = '^';
					en[y[i]] = 1;
				} else {
					en[x[i]] = en[y[i]];
					ans[i] = 'v';
				}
			}
			cout << ans << "\n";
			exit(0);
		}
	}

	cout << -1 << "\n";
}

int dp[500][500];
int go[M];

void solve2() {
	set<pair<int, int>> ss;
	for (int i = 0; i < m; ++i) {
		ss.insert({x[i], y[i]});
	}
	for (int i = 0; i < n; ++i) {
		for (int j = i + 1; j < n; ++j) {
			if (ss.count({i, j}) == 0) {
				for (int k = 0; k < m; ++k) {
					if (x[k] == i || x[k] == j) {
						ans[k] = '^';
					} else {
						ans[k] = 'v';
					}
				}
				cout << ans << "\n";
				exit(0);
			}
		}
	}


	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			dp[i][j] = -1;
		}
	}

	for (int i = 0; i < m; ++i) {
		for (int j = 0; j < n; ++j) {
			if (dp[y[i]][j] < M && dp[x[i]][j] == M) {
				dp[x[i]][j] = i;
				dp[j][x[i]] = i;
			}
			if (dp[x[i]][j] < M && dp[y[i]][j] == M) {
				dp[y[i]][j] = i;
				dp[j][y[i]] = i;
			}
		}
		go[i] = dp[x[i]][y[i]];
		dp[x[i]][y[i]] = dp[y[i]][x[i]] = M;
	}

	for (int a = 0; a < n; ++a) {
		for (int b = a + 1; b < n; ++b) {
			if (dp[a][b] < M) {
				for (int i = m - 1; i >= 0; --i) {
					dp[x[i]][y[i]] = dp[y[i]][x[i]] = go[i];
					for (int j = 0; j < n; ++j) {
						if (dp[y[i]][j] == i) {
							dp[y[i]][j] = M;
							dp[j][y[i]] = M;
						}
						if (dp[x[i]][j] == i) {
							dp[x[i]][j] = M;
							dp[j][x[i]] = M;
						}
					}
					if (x[i] == a && y[i] == b) {
						assert(false);
					}
					if (x[i] == b && y[i] == a) {
						assert(false);
					}
					if (x[i] == a) {
						ans[i] = '^';
						if (dp[a][b] == M) {
							a = y[i];
						}
					} else if (y[i] == a) {
						ans[i] = 'v';
						if (dp[a][b] == M) {
							a = x[i];
						}
					} else if (x[i] == b) {
						ans[i] = '^';
						if (dp[a][b] == M) {
							b = y[i];
						}
					} else if (y[i] == b) {
						ans[i] = 'v';
						if (dp[a][b] == M) {
							b = x[i];
						}
					} else {
						ans[i] = 'v';
					}
					assert(dp[a][b] < M);
				}
				cout << ans << "\n";
				exit(0);
			}
		}
	}

	cout << -1 << "\n";
}

int main() {
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cout.setf(ios::fixed), cout.precision(20);
	cin >> n >> m >> tt;
	for (int i = 0; i < m; ++i) {
		cin >> x[i] >> y[i];
		--x[i], --y[i];
	}

	if (tt == 1) {
		solve1();
	} else {
		solve2();
	}

	return 0;
}