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 <bits/stdc++.h>
#define fi first
#define se second
using namespace std;

const int N = 50005, M = 100005;

int n, m, t, u[M], v[M];
int x[M][2], y[M][2];
pair<bool, bool> tr[M][2];
bitset<N> bi[N];
bool chk[N];

bool trans(int &x, int &y, int u, int v, bool c) {
    if (x == u) {
        if (c) {
            x = v;
        }
        return !c;
    } else if (y == u) {
        if (c) {
            y = v;
        }
        return !c;
    } else if (x == v) {
        if (c) {
            x = u;
        }
        return c;
    } else {
        if (c) {
            y = u;
        }
        return c;
    }
}

void solve_1() {
    for (int i = 1; i <= n; i++) {
        bi[i].set(i);
    }
    for (int i = 1; i <= m; i++) {
        bi[v[i]] = (bi[u[i]] |= bi[v[i]]);
    }
    for (int i = 1; i <= n; i++) {
        if (bi[i].count() == n) {
            chk[i] = true;
            for (int j = m; j >= 1; j--) {
                if (chk[v[j]]) {
                    swap(u[j], v[j]);
                }
                chk[v[j]] |= chk[u[j]];
            }
            for (int j = 1; j <= m; j++) {
                cout << (u[j] < v[j] ? '^' : 'v');
            }
            return;
        }
    }
    cout << -1;
}

void solve_2() {
    if (n == 2) {
        return void(cout << -1);
    }
    if (u[1] == 1 && v[1] == 2) {
        x[0][0] = 1; y[0][0] = 3;
        x[0][1] = 2; y[0][1] = 3;
    } else if (u[1] == 1 && v[1] == 3) {
        x[0][0] = 1; y[0][0] = 2;
        x[0][1] = 2; y[0][1] = 3;
    } else {
        x[0][0] = 1; y[0][0] = 2;
        x[0][1] = 1; y[0][1] = 3;
    }
    for (int i = 1; i <= m; i++) {
        if (u[i] == x[i - 1][0] && v[i] == y[i - 1][0]) {
            for (int j = 0; j < 2; j++) {
                x[i][j] = x[i - 1][1]; y[i][j] = y[i - 1][1]; tr[i][j].fi = 1;
                tr[i][j].se = trans(x[i][j], y[i][j], u[i], v[i], j);
            }
        } else if (u[i] == x[i - 1][1] && v[i] == y[i - 1][1]) {
            for (int j = 0; j < 2; j++) {
                x[i][j] = x[i - 1][0]; y[i][j] = y[i - 1][0]; tr[i][j].fi = 0;
                tr[i][j].se = trans(x[i][j], y[i][j], u[i], v[i], j);
            }
        } else {
            for (int j = 0; j < 2; j++) {
                x[i][j] = x[i - 1][j]; y[i][j] = y[i - 1][j]; tr[i][j].fi = j;
                tr[i][j].se = trans(x[i][j], y[i][j], u[i], v[i], 0);
            }
        }
        for (int j = 0; j < 2; j++) {
            if (x[i][j] > y[i][j]) {
                swap(x[i][j], y[i][j]);
            }
        }
    }
    for (int i = m, cur = 0; i >= 1; i--) {
        chk[i] = tr[i][cur].se;
        cur = tr[i][cur].fi;
    }
    for (int i = 1; i <= m; i++) {
        cout << (chk[i] ? '^' : 'v');
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m >> t;
    for (int i = 1; i <= m; i++) {
        cin >> u[i] >> v[i];
    }
    if (t == 1) {
        solve_1();
    } else {
        solve_2();
    }
}