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>
//#pragma GCC optimize("Ofast")

#define fi first
#define se second

const int N = 100010;

using namespace std;

int n;
int m;
int T;
int x[N];
int y[N];
int s[N];
int res[N];
char f[N];
bitset < N / 2 > I[N / 2];

int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);

        ios_base::sync_with_stdio(0);

        cin >> n >> m >> T;
        for(int i = 1; i <= m; i++){
                cin >> x[i] >> y[i];
        }
        if(T == 2){
                if(n == 2){
                        cout << -1 << '\n';
                        return 0;
                }
                for(int i = 1; i <= n; i++){
                        res[i] = i;
                        s[i] = 1;
                }
                for(int i = m; i >= 1; i--){
                        int a = res[x[i]], b = res[y[i]];
                        if(s[a] < s[b]){
                                f[i] = '^';
                                s[a] += 1;
                                s[b] -= 1;
                                res[y[i]] = res[x[i]];
                        } else{
                                f[i] = 'v';
                                s[a] -= 1;
                                s[b] += 1;
                                res[x[i]] = res[y[i]];
                        }
                }
                for(int i = 1; i <= m; i++){
                        cout << f[i];
                }
                return 0;
        }
        for(int i = 1; i <= n; i++){
                I[i][i] = 1;
        }
        for(int i = m; i >= 1; i--){
                int a = x[i], b = y[i];
                I[a] |= I[b];
                I[b] = I[a];
        }
        for(int i = 2; i <= n; i++){
                I[1] &= I[i];
        }
        int x = -1;
        for(int i = 1; i <= n; i++){
                if(I[1][i]){
                        x = i;
                }
        }
        if(x == -1){
                cout << -1 << "\n";
                return 0;
        }
        res[x] = 1;
        for(int i = m; i >= 1; i--){
                int a = ::x[i], b = y[i];
                if(res[a]){
                        f[i] = '^';
                        res[b] = 1;
                } else{
                        f[i] = 'v';
                        res[a] |= res[b];
                }
        }
        for(int i = 1; i <= m; i++){
                cout << f[i];
        }
}