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:
#define _USE_MATH_DEFINES
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cstdio>
#include <complex>
#include <numeric>
#include <string.h>
#include <random>
#include <cmath>
#define rep(i,n) for (int i = 0; i < (int)n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
double PI = 3.141592653589793238462643383279;
const double EPS = 1e-9;
const ll MOD = 1000000007;
//const ll MOD = 998244353
const int inf = 1 << 30;
const ll linf = 1LL << 60;

int N, M, T;
int x[100000], y[100000];
ll b[50000];
int c[50000];
int c2[50000];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N >> M >> T;
    rep(i,M) cin >> x[i] >> y[i];
    rep(i,M) { x[i]--; y[i]--; }
    if (T == 1) {
        string ans;
        rep(i,M) ans.push_back('^');
        for (int i = 0; i < N; i += 64) {
            memset(b, 0, sizeof(b));
            for (int j = i; j < min(N,i+64); j++) {
                b[j] = (1LL<<(j-i));
            }
            for (int j = M-1; j >= 0; j--) {
                b[x[j]] |= b[y[j]];
                b[y[j]] |= b[x[j]];
            }
            for (int j = 1; j < N; j++) b[0] &= b[j];
            for (int j = 0; j < 64; j++) {
                if (!(b[0]&(1LL<<j))) continue;
                memset(b, 0, sizeof(b));
                b[j+i] = 1;
                for (int j = M-1; j >= 0; j--) {
                    if (b[x[j]] && !b[y[j]]) {
                        if (x[j] > y[j]) ans[j] = 'v';
                    }
                    if (b[y[j]] && !b[x[j]]) {
                        if (y[j] > x[j]) ans[j] = 'v';
                    }
                    b[x[j]] |= b[y[j]];
                    b[y[j]] |= b[x[j]];
                }
                cout << ans << endl;
                return 0;
            }
        }
        cout << -1 << endl;
    } else {
        if (N == 2) {
            cout << -1 << endl;
            return 0;
        }
        rep(i,N) c[i] = i;
        rep(i,N) c2[i] = 1;
        string ans;
        rep(i,M) ans.push_back('^');
        for (int i = M-1; i >= 0; i--) {
            if (c2[c[x[i]]] > 1) {
                c2[c[x[i]]]--;
                c2[c[y[i]]]++;
                c[x[i]] = c[y[i]];
                if (x[i] < y[i]) ans[i] = 'v';
            } else if (c2[c[y[i]]] > 0) {
                c2[c[x[i]]]++;
                c2[c[y[i]]]--;
                c[y[i]] = c[x[i]];
                if (x[i] > y[i]) ans[i] = 'v';
            } else if (c2[c[x[i]]] > 0) {
                c2[c[x[i]]]--;
                c2[c[y[i]]]++;
                c[x[i]] = c[y[i]];
                if (x[i] < y[i]) ans[i] = 'v';
            }
            /*
            rep(j,N) cerr << c2[j] << " ";
            cerr << endl;
            rep(j,N) cerr << c[j] << " ";
            cerr << endl;
            */
        }
        cout << ans << endl;
    }
}