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 <algorithm>
#include <bitset>
#include <cassert>
#include <iostream>
#include <vector>
using namespace std;

const int N_MAX = 5e4 + 5;

int N, M, T;
vector<int> X, Y;
vector<pair<int, int>> XY;

void solve_uniforming() {
    vector<bitset<N_MAX>> acceptable(N);

    for (int i = 0; i < N; i++)
        acceptable[i][i] = true;

    for (int i = M - 1; i >= 0; i--) {
        int x = X[i], y = Y[i];
        acceptable[x] = acceptable[y] = acceptable[x] | acceptable[y];
    }

    bitset<N_MAX> full;
    full.set();

    for (int i = 0; i < N; i++)
        full &= acceptable[i];

    if (full.none()) {
        cout << -1 << '\n';
        return;
    }

    int finish = -1;

    for (int i = 0; i < N; i++)
        if (full[i])
            finish = i;

    vector<bool> good(N, false);
    good[finish] = true;
    string answer(M, '.');

    for (int i = M - 1; i >= 0; i--) {
        int x = X[i], y = Y[i];

        if (!good[x])
            answer[i] = 'v';
        else
            answer[i] = '^';

        good[x] = good[y] = good[x] || good[y];
    }

    cout << answer << '\n';
}

void solve_non_uniforming() {
    if (N == 2) {
        cout << -1 << '\n';
        return;
    }

    string answer(M, '^');
    vector<int> relevant;

    for (int i = 0; i < M; i++)
        if (Y[i] < 3)
            relevant.push_back(i);

    if (relevant.empty()) {
        cout << answer << '\n';
        return;
    }

    int R = relevant.size();
    pair<int, int> current = {0, 1};

    if (current == XY[relevant.front()])
        current = {0, 2};

    for (int i = 0; i < R - 1; i++) {
        int index = relevant[i], next = relevant[i + 1];
        assert(current != XY[index]);

        pair<int, int> up = current;
        if (up.first == Y[index]) up.first = X[index];
        if (up.second == Y[index]) up.second = X[index];
        if (up.first > up.second) swap(up.first, up.second);

        pair<int, int> down = current;
        if (down.first == X[index]) down.first = Y[index];
        if (down.second == X[index]) down.second = Y[index];
        if (down.first > down.second) swap(down.first, down.second);

        if (up.first != up.second && up != XY[next]) {
            answer[index] = '^';
            current = up;
        } else {
            assert(down.first != down.second && down != XY[next]);
            answer[index] = 'v';
            current = down;
        }
    }

    cout << answer << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N >> M >> T;
    X.resize(M);
    Y.resize(M);
    XY.resize(M);

    for (int i = 0; i < M; i++) {
        cin >> X[i] >> Y[i];
        X[i]--; Y[i]--;
        XY[i] = {X[i], Y[i]};
    }

    if (T == 1)
        solve_uniforming();
    else
        solve_non_uniforming();
}