Problem:
Score : 1600 points

Problem Statement

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).

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.

Partial Score

800 points will be awarded for passing the testset satisfying T = 1.

800 points will be awarded for passing the testset satisfying T = 2.

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.

Sample Input 1

4 5 1
1 3
2 4
1 2
3 4
2 3

Sample Output 1

^^^^^

This state is uniforming: regardless of the starting wire, the token ends up at wire 1.

Sample Input 2

4 5 2
1 3
2 4
1 2
3 4
2 3

Sample Output 2

v^^^^

This state is non-uniforming: depending on the starting wire, the token might end up at wire 1 or wire 2.

Sample Input 3

3 1 1
1 2

Sample Output 3

-1

A uniforming state doesn't exist.

Sample Input 4

2 1 2
1 2

Sample Output 4

-1

A non-uniforming state doesn't exist.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = (int)(n) - 1; (i) >= (int)(m); -- (i))
#define ALL(x) std::begin(x), std::end(x)
using namespace std;

constexpr int N_MAX = 50000;

string solve1(int n, int m, const vector<int> & x, const vector<int> & y) {
    vector<bitset<N_MAX> > dp(n);
    REP (z, n) {
        dp[z][z] = true;
    }
    REP_R (i, m) {
        dp[x[i]] |= dp[y[i]];
        dp[y[i]] = dp[x[i]];
    }
    bitset<N_MAX> prod_dp;
    prod_dp.flip();
    REP (z, n) {
        prod_dp &= dp[z];
    }
    if (prod_dp.none()) {
        return "-1";
    }
    int goal = prod_dp._Find_first();
    string answer(m, '?');
    vector<bool> connected(m);
    connected[goal] = true;
    REP_R (i, m) {
        if (connected[x[i]]) {
            answer[i] = '^';
            connected[y[i]] = true;
        } else if (connected[y[i]]) {
            answer[i] = 'v';
            connected[x[i]] = true;
        } else {
            answer[i] = '^';
        }
    }
    return answer;
}

string solve2(int n, int m, const vector<int> & x, const vector<int> & y) {
    assert (false);
}

int main() {
    int n, m, t; cin >> n >> m >> t;
    vector<int> x(m), y(m);
    REP (i, m) {
        cin >> x[i] >> y[i];
        -- x[i];
        -- y[i];
    }
    string answer = (t == 1 ? solve1(n, m, x, y) : solve2(n, m, x, y));
    cout << answer << endl;
    return 0;
}