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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <stdio.h>
#include <time.h>
#include <numeric>
#include <random>
#include <unordered_map>
#include <unordered_set>
#define all(a) a.begin(),a.end()
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define pb push_back
#define debug(x) cerr << __LINE__ << ' ' << #x << ':' << x << '\n'
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> P;
typedef complex<ld> com;
constexpr int inf = 1000000010;
constexpr ll INF = 1000000000000000010;
constexpr ld eps = 1e-12;
constexpr ld pi = 3.141592653589793238;
template<class T, class U> inline bool chmax(T &a, const U &b) { if (a < b) { a = b; return true; } return false; }
template<class T, class U> inline bool chmin(T &a, const U &b) { if (a > b) { a = b; return true; } return false; }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);

	ll n, m, t;
	cin >> n >> m >> t;
	vector<P> p(m);
	rep(i, m) {
		cin >> p[i].first >> p[i].second;
		p[i].first--; p[i].second--;
	}
	if (t == 1) {
		vector<bitset<50010>> bst(n);
		rep(i, n) bst[i][i] = 1;
		rep(i, m) {
			bst[p[i].first] |= bst[p[i].second];
			bst[p[i].second] |= bst[p[i].first];
		}
		int x = -1;
		rep(i, n) if (bst[i].count() == n) {
			x = i;
			break;
		}
		if (x == -1) {
			cout << "-1\n";
			return 0;
		}
		vector<bool> ok(n);
		ok[x] = true;
		string ans = "";
		for (int i = m - 1; i >= 0; i--) {
			if (ok[p[i].first] && !ok[p[i].second]) {
				ans += '^';
				ok[p[i].second] = true;
			}
			else if (!ok[p[i].first] && ok[p[i].second]) {
				ans += 'v';
				ok[p[i].first] = true;
			}
			else {
				ans += '^';
			}
		}
		reverse(all(ans));
		cout << ans << '\n';
	}
	if (t == 2) {
		set<P> st;
		rep(i, m) st.insert({ p[i].first,p[i].second });
		int x = -1, y = -1;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (!st.count({ i,j })) {
					x = i, y = j;
					i = n, j = n;
				}
			}
		}
		if (x != -1 && y != -1) {
			string ans = "";
			rep(i, m) {
				if (p[i].second == x || p[i].second == y) ans += 'v';
				else ans += '^';
			}
			cout << ans << '\n';
		}
		else {
			chmin(n, 3);
			vector<vector<bitset<320>>> bst(m + 1, vector<bitset<320>>(n));
			rep(i, n) {
				rep(j, n) {
					if (i != j) bst[0][i][j] = 1;
				}
			}
			rep(i, m) {
				rep(j, n) {
					bst[i + 1][j] = bst[i][j];
				}
				int pf = p[i].first; int ps = p[i].second;
				if (pf >= n || ps >= n) continue;
				bst[i + 1][pf][ps] = 0;
				bst[i + 1][ps][pf] = 0;
				bst[i + 1][pf] |= bst[i + 1][ps];
				bst[i + 1][ps] |= bst[i + 1][pf];
				rep(j, n) {
					if (j == pf || j == ps) continue;
					if (bst[i + 1][pf][j]) bst[i + 1][j][pf] = 1;
					if (bst[i + 1][ps][j]) bst[i + 1][j][ps] = 1;
				}
			}
			int x = -1, y = -1;
			rep(i, n) {
				rep(j, n) {
					if (bst[m][i][j]) {
						x = i, y = j;
					}
				}
			}
			if (x == -1 && y == -1) {
				cout << "-1\n";
				return 0;
			}
			string ans = "";
			for (int i = m - 1; i >= 0; i--) {
				if (bst[i][x][y]) {
					if (x == p[i].first || y == p[i].first) ans += '^';
					else ans += 'v';
				}
				else {
					if (x == p[i].first) {
						x = p[i].second;
						ans += '^';
					}
					else if (y == p[i].first) {
						y = p[i].second;
						ans += '^';
					}
					else if (x == p[i].second) {
						x = p[i].first;
						ans += 'v';
					}
					else if (y == p[i].second) {
						y = p[i].first;
						ans += 'v';
					}
				}
			}
			reverse(all(ans));
			cout << ans << '\n';
		}
	}
}