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
#include <bits/stdc++.h>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

constexpr ll mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;

template <typename T> constexpr bool chmax(T &a, const T b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

string solve1(int n, vector<int> x, vector<int> y) {
    int m = x.size();
    vector<bitset<50009>> a(n);
    rep(i, n) a[i].set(i);
    rep(i, m) a[x[i]] = a[y[i]] = a[x[i]] | a[y[i]];
    rep(i, n) {
        if(int(a[i].count()) != n) continue;
        bitset<50009> b;
        b.set(i);
        string ret(m, 'v');
        while(m--) {
            if(b[x[m]]) {
                b.set(y[m]);
                ret[m] = '^';
            } else
                b[x[m]] = b[y[m]];
        }
        return ret;
    }
    return "-1";
}

string solve2(int n, vector<int> x, vector<int> y) {
    if(n == 2) return "-1";
    int m = x.size();
    string ret(m, '^');
    vector<int> a(n), num(n);
    rep(i, n) {
        a[i] = i;
        num[i] = 1;
    }
    for(int i = m - 1; i >= 0; i--) {
        if(num[a[x[i]]] > 1) {
            num[a[x[i]]]--;
            num[a[y[i]]]++;
            a[x[i]] = a[y[i]];
            ret[i] = 'v';
        } else {
            num[a[x[i]]]++;
            num[a[y[i]]]--;
            a[y[i]] = a[x[i]];
        }
    }
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int n, m, t;
    cin >> n >> m >> t;

    vector<int> x(m), y(m);
    rep(i, m) {
        cin >> x[i] >> y[i];
        x[i]--;
        y[i]--;
    }
    if(t == 1) cout << solve1(n, x, y) << endl;
    if(t == 2) cout << solve2(n, x, y) << endl;

    return 0;
}