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:
#define _CRT_SECURE_NO_WARNINGS
#define _SCL_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <utility>
#include <algorithm>
#include <functional>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <limits>
#include <numeric>
#include <valarray>
#include <fstream>
#include <array>
#include <random>
#include <unordered_set>
#include <unordered_map>

using namespace std;
using uint = uint32_t;
using LL = int64_t;
using ULL = uint64_t;
using PP = pair<LL, LL>;
template <typename T> using PriorityQ = priority_queue<T, vector<T>, greater<T> >;
#define REP(i, a, n) for(LL i = (a), i##_max_ = (n); i < i##_max_; ++i)
#define REM(i, a, n) for(LL i = (LL)(n) - 1, i##_min_ = (a); i >= i##_min_; --i)
#define FLOAT fixed << setprecision(16)
#define SPEEDUP { cin.tie(NULL); ios::sync_with_stdio(false); }
const int INF = 0x3FFFFFFF;
const LL INFLL = 0x3FFFFFFF3FFFFFFF;
const double INFD = 1.0e+308;
const double EPS = 1.0e-9;

void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
template <class T, class U> istream& operator>>(istream& ist, pair<T, U>& right) { return ist >> right.first >> right.second; }
template <class T, class U> ostream& operator<<(ostream& ost, const pair<T, U>& right) { return ost << right.first << ' ' << right.second; }
template <class T, class TCompatible, size_t N> void Fill(T(&dest)[N], const TCompatible& val) { fill(dest, dest + N, val); }
template <class T, class TCompatible, size_t M, size_t N> void Fill(T(&dest)[M][N], const TCompatible& val) { for (int i = 0; i < M; ++i) Fill(dest[i], val); }
template <class T> T Next() { T buf; cin >> buf; return buf; }
istream& Ignore(istream& ist) { string s; ist >> s; return ist; }
bool Inside(int i, int j, int h, int w) { return i >= 0 && i < h && j >= 0 && j < w; }

#ifdef ONLY_MY_ENVIR
#include "IntMod.h"
#include "Algebraic.h"
#include "BinaryMatrix.h"
#include "BinaryTree.h"
#include "Bipartite.h"
#include "BIT.h"
#include "DynamicMod.h"
#include "Factorization.h"
#include "FFT.h"
#include "FlowSolver.h"
#include "Graph.h"
#include "LazySegmentTree.h"
#include "Math.h"
#include "Matrix.h"
#include "MinCostFlowSolver.h"
#include "MinMax.h"
#include "Numbers.h"
#include "Polynomial.h"
#include "Position.h"
#include "Range.h"
#include "Rational.h"
#include "SegmentTree.h"
#include "SegmentTree2D.h"
#include "SlidingWindow.h"
#include "SuffixArray.h"
#include "Tree.h"
#include "UnionFind.h"
#endif

#ifdef __GNUC__
typedef __int128 LLL;
istream& operator>>(istream& ist, __int128& val) { LL tmp; ist >> tmp; val = tmp; return ist; }
ostream& operator<<(ostream& ost, __int128 val) { LL tmp = val; ost << tmp; return ost; }
#endif

void TLE() {
	const int BIG = 1e6;
	volatile int sum = 0;
	for (int i = 0; i < BIG; ++i) {
		for (int j = 0; j < BIG; ++j) {
			sum += i * j;
		}
	}
}

void RE() {
	throw 1;
}

void MLE() {
	std::vector<int> big_array(1 << 29);
	std::cerr << big_array.back() << std::endl;
	exit(0);
}

void OLE() {
	for (int i = 0; i < (1 << 24); ++i) {
		printf("AtCoderAtCoderAtCoderAtCoder\n");
	}
	exit(0);
}

using Bits = bitset<50000>;
int N, M, T;
int X[100000], Y[100000];
Bits bits[50000];
bool used[50000];

map<PP, vector<int>> mp;
int get(PP p, int j) {
	if (p.first > p.second) swap(p.first, p.second);
	if (!mp.count(p)) return INF;
	const auto& v = mp[p];
	auto it = lower_bound(v.begin(), v.end(), j);
	if (it == v.end()) return INF;
	return *it;
}

void probB() {
	if (N == 2) {
		cout << -1 << endl;
		return;
	}

	REP(j, 0, M) {
		mp[PP(X[j], Y[j])].push_back(j);
	}

	int x = 0;
	int y = 1;
	if (get(PP(0, 1), 0) < get(PP(0, 2), 0)) {
		y = 2;
	}

	string s(M, 'v');
	REP(j, 0, M) {
		int a = X[j];
		int b = Y[j];

		bool end = a == x && b == y;
		assert(!end);

		if (x == a) {
			PP p(a, y);
			PP q(b, y);
			if (get(p, j) < get(q, j)) {
				s[j] = 'v';
				x = b;
			} else {
				s[j] = '^';
			}
		} else if (x == b) {
			PP p(a, y);
			PP q(b, y);
			if (get(p, j) < get(q, j)) {
				s[j] = 'v';
			} else {
				s[j] = '^';
				x = a;
			}
		} else if (y == a) {
			PP p(x, a);
			PP q(x, b);
			if (get(p, j) < get(q, j)) {
				s[j] = 'v';
				y = b;
			} else {
				s[j] = '^';
			}
		} else if (y == b) {
			PP p(x, a);
			PP q(x, b);
			if (get(p, j) < get(q, j)) {
				s[j] = 'v';
			} else {
				s[j] = '^';
				y = a;
			}
		}
		if (x > y) swap(x, y);
	}
	cout << s << endl;
}

void subA(int from) {
	string s(M, 'v');
	used[from] = true;
	REM(j, 0, M) {
		int a = X[j];
		int b = Y[j];

		if (used[a]) {
			s[j] = '^';
			used[b] = true;
		} else if (used[b]) {
			used[a] = true;
		}
	}
	cout << s << endl;
}

void probA() {
	REP(i, 0, N) {
		bits[i][i] = true;
	}

	REP(j, 0, M) {
		int a = X[j];
		int b = Y[j];
		bits[a] |= bits[b];
		bits[b] |= bits[a];
	}

	REP(i, 0, N) {
		if (bits[i].count() == N) {
			subA(i);
			return;
		}
	}
	cout << -1 << endl;
}

int main() {
	cin >> N >> M >> T;
	REP(j, 0, M) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		X[j] = a;
		Y[j] = b;
	}
	
	if (T == 1) {
		probA();
	} else {
		probB();
	}

	return 0;
}