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
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
  
#include <iostream>
#include <stdlib.h>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <assert.h>
#include <memory.h>
#include <time.h>
#include <bitset>
  
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define ld long double
#define rep(i, l, r) for (int i = (l); i < (r); i++)
#define repb(i, r, l) for (int i = (r); i > (l); i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define mp(a, b) make_pair(a, b)
#define rank qwertyuio
#define next dfghjk
#define prev fhsgfhjf
#define plus fsghsf
#define minus ytryr
  
using namespace std;
  
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(ll &x, ll y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(ll &x, ll y) { return (y > x) ? x = y, 1 : 0; }
  
const int N = 50000;
const int inf = (int)1e9 + 1;
const ll big = (ll)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const ld eps = 1e-12;
const double pi = atan2(0, -1);
const int ABC = 26;

bitset<N> mask[N];

void solve1(int n, int m) {
    rep(i, 0, n) {
        mask[i][i] = 1;
    }
    vector<pair<int, int> > a;
    rep(i, 0, m) {
        int x, y;
        cin >> x >> y; x--; y--;
        mask[x] |= mask[y];
        mask[y] = mask[x];
        a.push_back({x, y});
    }
    int r = -1;
    rep(i, 0, n) {
        if (mask[i].count() == n) {
            r = i;
            break;
        }
    }
    if (r == -1) {
        cout << -1 << "\n";
        return;
    }
    bool used[n];
    fill(used, used + n, false);
    used[r] = true;
    string ans(m, ' ');
    repb(i, m - 1, -1) {
        int x = a[i].fi, y = a[i].se;
        if (used[x]) {
            ans[i] = '^';
            used[y] = true;
        } else if (used[y]) {
            ans[i] = 'v';
            used[x] = true;
        } else {
            ans[i] = '^';
        }
    }
    cout << ans << "\n";
}

void solve2(int n, int m) {
    if (n == 2) {
        cout << -1 << "\n";
        return;
    }
    vector<pair<int, int> > a;
    rep(i, 0, m) {
        int x, y;
        cin >> x >> y; x--; y--;
        a.push_back({x, y});
    }
    bool used[n];
    fill(used, used + n, false);
    used[0] = true; int cur = 1;
    string ans(m, ' ');
    repb(i, m - 1, -1) {
        int x = a[i].fi, y = a[i].se;
        if (cur == 1) {
            if (used[x]) {
                ans[i] = '^';
                used[y] = true;
                cur = 2;
            } else if (used[y]) {
                ans[i] = 'v';
                used[x] = true;
                cur = 2;
            } else {
                ans[i] = '^';
            }
        } else {
            if ((used[x] && used[y]) || (!used[x] && !used[y])) {
                ans[i] = '^';
            } else if (used[x]) {
                ans[i] = 'v';
                used[x] = false;
                cur = 1;
            } else {
                ans[i] = '^';
                used[y] = false;
                cur = 1;
            }
        }
    }
    cout << ans << "\n";
}

int main()
{
    //freopen("a.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    //ll TL = 10.95 * CLOCKS_PER_SEC;
    //clock_t time = clock();
    int n, m, T;
    cin >> n >> m >> T;
    if (T == 1) {
        solve1(n, m);
    } else {
        solve2(n, m);
    }
    return 0;
}