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 long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n, m, t;
    cin >> n >> m >> t;
    vector<int> x(m), y(m);
    for (int i = 0; i < m; i++) {
        cin >> x[i] >> y[i];
        x[i]--; y[i]--;
    }
    if (t == 2) {
        if (n == 2) {
            cout << -1;
            return 0;
        }
        int b = x[0];
        int c = y[0];
        int k = 0;
        int a = 0;
        string s(m, 'v');
        while (a == b || a == c) a++;
        for (int i = 1; i < m; i++) {
            if (x[i] == b && y[i] == c) {
                k = i;
            } else if (x[i] == a || y[i] == a) {
                int xx = x[i];
                int yy = y[i];
                if (yy == a) swap(xx, yy);
                if (yy != b && yy != c) {
                    if (x[i] == a) {
                        s[i] = '^';
                    } else {
                        s[i] = 'v';
                    }
                } else {
                    int aa, bb, cc;
                    if (yy == b) {
                        s[k] = 'v';
                        aa = c;
                        if (a < b) {
                            bb = a;
                            cc = b;
                        } else {
                            bb = b;
                            cc = a;
                        }
                    } else {
                        s[k] = '^';
                        aa = b;
                        if (a < c) {
                            bb = a;
                            cc = c;
                        } else {
                            bb = c;
                            cc = a;
                        }
                    }
                    a = aa;
                    b = bb;
                    c = cc;
                    k = i;
                }
            } else if (x[i] == b || x[i] == c) {
                s[i] = '^';
            }
        }
        cout << s;
    } else {
        const int MAX = 50000;
        vector<bitset<MAX>> a(n);
        for (int i = 0; i < n; i++) a[i][i] = 1;
        for (int i = m - 1; i >= 0; i--) {
            a[y[i]] |= a[x[i]];
            a[x[i]] = a[y[i]];
        }
        bitset<MAX> xx;
        for (int i = 0; i < n; i++) {
            xx[i] = 1;
        }
        for (int i = 0; i < n; i++) {
            xx &= a[i];
        }
        int t = -1;
        for (int i = 0; i < n; i++) {
            if (xx[i]) t = i;
        }
//        cerr << t << "\n";
        if (t == -1) {
            cout << -1;
        } else {
            string s(m, '.');
            bitset<MAX> z;
            z[t] = 1;
            for (int i = m - 1; i >= 0; i--) {
                if (z[x[i]]) {
                    s[i] = '^';
                    z[y[i]] = 1;
                }
                else if (z[y[i]]) {
                    s[i] = 'v';
                    z[x[i]] = 1;
                } else {
                    s[i] = 'v';
                }
            }
            cout << s << "\n";
        }
    }

    return 0;
}