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 <cstring>
#include <algorithm>
using namespace std;

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	int N, M, T;
	cin >> N >> M >> T;
	if(T == 1) {
		vector< vector<uint64_t> > bits(50000, vector<uint64_t>(50000/64+1, 0));
		for(int i = 0; i < N; i++) bits[i][i>>6] |= 1ULL<<(i&63);
		for(int j = 0; j < N; j++)
			for(int i = N; i < (N/64+1)*64; i++) bits[j][i>>6] |= 1ULL<<(i&63);
		vector< pair<int, int> > bal(M);
		int final = -1;
		for(int i = 0; i < M; i++) {
			int x, y;
			cin >> x >> y;
			bal[i] = {--x, --y};
			if(final != -1) continue;
			uint64_t * __restrict ptr_x = &(bits[x][0]);
			uint64_t * __restrict ptr_y = &(bits[y][0]);
			uint64_t * __restrict ptr_fin = ptr_x + (N/64+1);
			uint64_t done = -1;
			for(; ptr_x != ptr_fin; ptr_x++, ptr_y++) {
				uint64_t v = *ptr_x | *ptr_y;
				*ptr_x = *ptr_y = v;
				done &= v;
			}
			if(!(~done)) final = x;
		}
		if(final == -1) {
			cout << "-1\n";
			return 0;
		}
		vector<int> max_id(N, -1);
		max_id[final] = M;
		string ans;
		for(int i = M-1; i >= 0; i--) {
			int x = bal[i].first, y = bal[i].second;
			if(max_id[x] == -1 && max_id[y] == -1)
				ans += '^';
			else if(max_id[x] != -1 && max_id[y] != -1)
				ans += (max_id[x] > max_id[y]) ? '^' : 'v';
			else if(max_id[x] == -1) {
				ans += 'v';
				max_id[x] = i;
			}
			else {
				ans += '^';
				max_id[y] = i;
			}
		}
		reverse(begin(ans), end(ans));
		cout << ans << "\n";
		return 0;
	}
	// T == 2
	vector< pair<int, int> > bal(M);
	for(int i = 0; i < M; i++) {
		int x, y;
		cin >> x >> y;
		bal[i] = {--x, --y};
	}
	if(1LL*N*(N-1)/2 > M) {
		vector<int> cnt_bal(N, 0);
		for(int i = 0; i < M; i++) {
			cnt_bal[bal[i].first]++;
			cnt_bal[bal[i].second]++;
		}
		for(int i = 0; i < N; i++) if(cnt_bal[i] < N-1) {
			vector<char> adj(N, 0);
			adj[i] = 1;
			for(int j = 0; j < M; j++) {
				if(bal[j].first == i) adj[bal[j].second] = 1;
				if(bal[j].second == i) adj[bal[j].first] = 1;
			}
			for(int j = 0; j < N; j++) if(!adj[j]) {
				string ans;
				for(int k = 0; k < M; k++) {
					if(bal[k].first == i || bal[k].first == j) ans += '^';
					else ans += 'v';
				}
				cout << ans << "\n";
				return 0;
			}
			break;
		}
		return 0;
	}
	vector< vector< pair<int, int> > > pair_add(M), pair_rem(M);
	vector< vector<char> > pair_live(N, vector<char>(N, 1));
	for(int i = M-1; i >= 0; i--) {
		int x = bal[i].first, y = bal[i].second;
		for(int j = 0; j < N; j++) if(j != x && j != y && pair_live[x][j]^pair_live[y][j]) {
			if(!pair_live[x][j] && pair_live[y][j]) pair_add[i].push_back({x, j});
			if(!pair_live[y][j] && pair_live[x][j]) pair_add[i].push_back({y, j});
		}
		sort(begin(pair_add[i]), end(pair_add[i]));
		auto it = unique(begin(pair_add[i]), end(pair_add[i]));
		pair_add[i].resize(it-begin(pair_add[i]));
		for(auto p : pair_add[i])
			pair_live[p.first][p.second] = pair_live[p.second][p.first] = 1;
		if(pair_live[x][y]) {
			pair_rem[i] = {{x, y}};
			pair_live[x][y] = pair_live[y][x] = 0;
		}
	}
	int l = -1, r = -1;
	for(int i = 0; i < N; i++) for(int j = 0; j < i; j++) if(pair_live[i][j]) {
		l = i, r = j;
		break;
	}
	if(l == -1) {
		cout << "-1\n";
		return 0;
	}
	string ans;
	for(int i = 0; i < M; i++) {
		for(auto p : pair_rem[i])
			pair_live[p.first][p.second] = pair_live[p.second][p.first] = 1;
		for(auto p : pair_add[i])
			pair_live[p.first][p.second] = pair_live[p.second][p.first] = 0;
		int x = bal[i].first, y = bal[i].second;
		if(l == x && r == y) return 0;
		if(l == y && r == x) return 0;
		if(l != x && l != y && r != x && r != y) {
			ans += '^';
			continue;
		}
		if(l == x && pair_live[y][r]) {
			ans += 'v';
			l = y;
		}
		else if(l == y && pair_live[x][r]) {
			ans += '^';
			l = x;
		}
		else if(r == x && pair_live[l][y]) {
			ans += 'v';
			r = y;
		}
		else if(r == y && pair_live[l][x]) {
			ans += '^';
			r = x;
		}
		else {
			if(!pair_live[l][r]) return 0;
			if(l == y || r == y) ans += 'v';
			else ans += '^';
		}
	}
	cout << ans << "\n";
}