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>

using namespace std;

#define ll  long long
#define ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 5e4;

typedef pair<int,int>   ii;
typedef bitset<N>       bs;

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

    int n;  cin >> n;
    int m;  cin >> m;
    int T;  cin >> T;

    vector<ii>  E;

    for(int i = 0 ; i < m ; ++i)    {
        int x;  cin >> x;
        int y;  cin >> y;

        E.pb(x,y);
    }

    reverse(all(E));

    if (T == 1) {
        vector<bs>  f(n,0);
        bs  S;

        for(int i = 0 ; i < n ; ++i)    f[i][i] = S[i] = 1;
        for(int i = 0 ; i < m ; ++i)    {
            int x = E[i].X - 1;
            int y = E[i].Y - 1;

            f[x] = f[y] = (f[x] | f[y]);
        }

        for(int i = 0 ; i < n ; ++i)
            S &= f[i];

        int x = n;

        for(int i = 0 ; i < n ; ++i)
            if (S[i])   {
                x = i;
                break;
            }

        if (x < n)  {
            vector<int> t(n,0); t[x] = 1;

            string res = "";

            for(int i = 0 ; i < m ; ++i)    {
                int x = E[i].X - 1;
                int y = E[i].Y - 1;

                if (t[x])   res += '^', t[y] = 1;
                else        res += 'v', t[x] = t[y];
            }
            reverse(all(res));  cout << res << endl;
        }
        else    cout << "-1";
    }
    if (T == 2) {
        if (n < 3)  return  cout << "-1",0;

        vector<int> f(n);
        vector<int> c(n);

        iota(all(f),0);
        fill(all(c),1);

        string res = "";

        for(int i = 0 ; i < m ; ++i)    {
            int x = E[i].X - 1;
            int y = E[i].Y - 1;

            if (c[f[x]] > 1)
                c[f[x]]--,
                c[f[y]]++,
                f[x] = f[y],
                res += 'v';
            else
                c[f[y]]--,
                c[f[x]]++,
                f[y] = f[x],
                res += '^';
        }
        reverse(all(res));  cout << res << endl;
    }
}