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;
typedef long long ll;
typedef long double ld;
const int maxn = 5e4 + 10;

bitset<maxn> bs[maxn];
int x[2*maxn], y[2*maxn];
int par[maxn];
bool dir[2*maxn];
int b[maxn], cnt[maxn];

int get_par(int v){ return par[v] < 0 ? v : par[v] = get_par(par[v]); }

void merge(int v, int u){
	if ((v = get_par(v)) == (u = get_par(u)))
		return;
	par[v] = u;
}

int main(){
	ios_base::sync_with_stdio(false);
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	int n, m, T;
	cin >> n >> m >> T;
	for (int i = 1; i <= m; i++)
		cin >> x[i] >> y[i];
	if (T == 1){
		for (int i = 1; i <= n; i++)
			bs[i][i] = 1;
		for (int i = 1; i <= m; i++){
			bs[x[i]] |= bs[y[i]];
			bs[y[i]] |= bs[x[i]];
		}
		int root = -1;
		for (int i = 1; i <= n; i++)
			if (bs[i].count() == n)
				root = i;
		if (root == -1)
			return cout << -1 << endl, 0;
		memset(par, -1, sizeof par);
		string s;
		for (int i = m; i >= 1; i--){
			if (get_par(x[i]) == get_par(root)){
				s += '^';
				merge(x[i], y[i]);
			}
			else if (get_par(y[i]) == get_par(root)){
				s += 'v';
				merge(x[i], y[i]);
			}
			else
				s += 'v';
		}
		reverse(s.begin(), s.end());
		cout << s << endl;
	}
	else{
		if (n == 2)
			return cout << -1 << endl, 0;
		for (int i = 1; i <= n; i++)
			b[i] = i, cnt[i] ++;
		string s;
		for (int i = m; i >= 1; i--){
			if (cnt[b[x[i]]] >= 2){
				s += 'v';
				cnt[b[x[i]]] --, cnt[b[y[i]]] ++;
				b[x[i]] = b[y[i]];
			}
			else if (cnt[b[y[i]]] >= 2){
				s += '^';
				cnt[b[x[i]]] ++, cnt[b[y[i]]] --;
				b[y[i]] = b[x[i]];
			}
			else{
				s += '^';
				cnt[b[x[i]]] ++, cnt[b[y[i]]] --;
				b[y[i]] = b[x[i]];
			}
		}
		reverse(s.begin(), s.end());
		cout << s << endl;
	}
}