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;
 
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;
	int T;
	cin >> T;
	vector<int> x(m);
	vector<int> y(m);
	for(int i = 0; i < m; i++){
		cin >> x[i] >> y[i];
		x[i]--; y[i]--;
	}
	if(T == 1){
		using B = bitset<50000>;
		vector<B> cur(n);
		for(int i = 0; i < n; i++){
			cur[i][i].flip();
		}
		for(int i = 0; i < m; i++){
			cur[x[i]] |= cur[y[i]];
			cur[y[i]] |= cur[x[i]];
		}
		for(int i = 0; i < n; i++){
			if((int)cur[i].count() == n){
				vector<int> where(n, 0);
				where[i] = 1;
				string st(m, '.');
				for(int j = m-1; j >= 0; j--){
					st[j] = "^v"[where[x[j]] < where[y[j]]];
					where[x[j]] |= where[y[j]];
					where[y[j]] |= where[x[j]];
				}
				cout << st << '\n';
				return 0;
			}
		}
		cout << -1 << '\n';
	} else if(T == 2){
		if(n == 2){
			cout << -1 << '\n';
			return 0;
		}
		string st(m, '^');
		vector<int> reach(3, 1);
		int last = -1;
		for(int i = 0; i < m; i++){
			if(y[i] >= 3) continue;
			int other = 3 - x[i] - y[i];
			if(!reach[other]){
				assert(last >= 0);
				st[last] ^= ('^' ^ 'v');
				reach[other] = true;
			}
			st[i] = '^';
			reach[x[i]] = true;
			reach[y[i]] = false;
			last = i;
		}
		cout << st << '\n';
	}
}