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>
using namespace std;
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

typedef bitset<50000> bs;

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

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

	vi x(m), y(m);
	rep(i,0,m) cin >> x[i] >> y[i];
	trav(xx, x) --xx;
	trav(xx, y) --xx;

	if(t == 2){
		if(n == 2){
			cout << -1 << endl;
			return 0;
		} else {
			vector<stack<int>> ss(n);
			rep(i,0,n) ss[i].push(-1);
			rrep(i,0,m){
				ss[x[i]].push(y[i]);
				ss[y[i]].push(x[i]);
			}
			int a = 0;
			int b = ss[a].top() == 1 ? 2 : 1;
			string ans;
			rep(i,0,m){
				ss[x[i]].pop();
				ss[y[i]].pop();
				if(x[i] == a || x[i] == b){
					if(x[i] == b) swap(a,b);
					if(ss[b].top() == a){
						ans.push_back('v');
						a = y[i];	
					} else {
						ans.push_back('^');	
					}
				} else if(y[i] == a || y[i] == b){
					if(y[i] == b) swap(a,b);
					if(ss[b].top() == a){
						ans.push_back('^');
						a = x[i];	
					} else {
						ans.push_back('v');	
					}
				} else {
					ans.push_back('v');
				}
			}
			cout << ans << endl;
			return 0;
		}
	}

	vector<bs> cover(n, 0);
	rep(i,0,n) cover[i].set(i);

	rep(i,0,m){
		cover[x[i]] = cover[y[i]] |= cover[x[i]];
	}
	
	rep(i,0,n){
		if(int(cover[i].count()) == n){
			string ans(m, 'v');
			bs good(0);
			good.set(i);
			rrep(i,0,m){
				if(good[x[i]]){
					good.set(y[i]);
					ans[i] = '^';
				} else if(good[y[i]]){
					good.set(x[i]);
					ans[i] = 'v';
				}
			}
			cout << ans << endl;
			return 0;
		}
	}
	cout << "-1" << endl;
}
