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 pb push_back
#define pll pair <ll, ll>
#define mp make_pair
#define pyshnapyshnakaa ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define x first
#define y second
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
#define plll pair <pair <ll, ll>, ll>
#define pllll pair <pair <ll, ll>, pair <ll, ll> >
#define psl pair <string, ll>
#define pss pair <string, string>
#define pssss pair <pss, pss>
#define all(a) a.begin(), a.end()

typedef long long ll;

typedef long double ld;

using namespace std;

const ll maxn = 1e6;
const ll maxn1 = 5e4 + 100;
const ll inf = 1e16;

ll n, m, k, t;

struct edge {
    ll to;
    ll i;
};

vector <edge> V[maxn];

vector <pll> E;

// set <ll> S[maxn];

bool ANS[maxn];

ll B[maxn];

ll cnt[maxn];

bitset <50000> X[maxn1];

bitset <50000> X2[maxn1];

int main() {
    pyshnapyshnakaa;
    ll q, w, e, a, b, c;
    cin >> n >> m >> t;
    for (q = 0; q < m; q++) {
        cin >> a >> b;
        a--; b--;
        E.pb(mp(a, b));
        // S[a].insert(b);
        // S[b].insert(a);
        edge e, f;
        e.to = b; f.to = a;
        e.i = f.i = q;
        V[a].pb(e);
        V[b].pb(f);
    }
    if (t == 2) {
        if (n == 2 && m >= 1) {
            cout << -1;
            return 0;
        }
        for (q = 0; q < n; q++) {
            B[q] = q;
            cnt[q] = 1;
        }
        for (w = E.size() - 1; w >= 0; w--) {
            ll a = E[w].first, b = E[w].second;
            if (a > b) {
                swap(a, b);
            }
            if (cnt[B[a]] >= 2 || cnt[B[a]] + cnt[B[b]] != n) {
                ANS[w] = 1;
                cnt[B[a]]--;
                B[a] = B[b];
                cnt[B[a]]++;
            }
            else {
                ANS[w] = 0;
                cnt[B[b]]--;
                B[b] = B[a];
                cnt[B[b]]++;   
            }
        }
        for (q = 0; q < m; q++) {
            if (ANS[q]) {
                cout << "v";
            }
            else {
                cout << "^";
            }
        }
    }
    else {
        for (q = 0; q < n; q++) {
            X[q][q] = 1;
            X2[q][q] = 1;
        }
        for (w = E.size() - 1; w >= 0; w--) {
            ll a = E[w].first, b = E[w].second;
            X[a] |= X[b];
            X[b] = X[a];
        }
        ll ansv = -1;
        bitset <50000> X1;
        for (q = 0; q < 50000; q++) {
            X1[q] = 1;
        }
        for (q = 0; q < n; q++) {
            X1 &= X[q];
        }
        for (q = 0; q < n; q++) {
            if (X1[q] == 1) {
                ansv = q;
            }
        }
        if (ansv == -1) {
            cout << -1;
            return 0;
        }
        for (w = E.size() - 1; w >= 0; w--) {
            ll a = E[w].first, b = E[w].second;
            if (b < a) {
                swap(a, b);
            }
            if (X2[a][ansv] == 0 && X2[b][ansv] == 1) {
                ANS[w] = 1;
            }
            X2[a] |= X2[b];
            X2[b] = X2[a];
        }
        for (q = 0; q < m; q++) {
            if (ANS[q]) {
                cout << "v";
            }
            else {
                cout << "^";
            }
        }
    }
    return 0;
}