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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
typedef vector<vector<ll>> mat;
typedef vector<ll> vec;

void debug(vec &v) {
	int n = v.size();
	rep(i, n) {
		if (i > 0)cout << " ";
		cout << v[i];
	}
	cout << endl;
}

int n, m, t;

bool used[1 << 16];
bitset<50000> bt[50000];
bitset<50000> al;
bitset<50000> ori;
void solve1() {
	vector<int> a(m), b(m);
	rep(i, m) {
		cin >> a[i] >> b[i]; a[i]--; b[i]--;
	}
	string ans; ans.resize(m);
	rep(i, n)bt[i][i] = 1;
	per(j, m) {
		bt[a[j]] |= bt[b[j]];
		bt[b[j]] = bt[a[j]];
	}
	rep(i, n)al[i] = 1;
	rep(i, n) {
		al &= bt[i];
	}

	rep(i, n) {
		if (al[i]) {
			ori = 0; ori[i] = true;
			per(j, m) {
				if (ori[a[j]])ans[j] = '^';
				else ans[j] = 'v';
				ori[a[j]] = ori[b[j]] = ori[a[j]] | ori[b[j]];
			}
			cout << ans << endl; return;
		}
	}
	cout << -1 << endl;
}

vector<string> s3 = { "vv^","vv^","v^v" };
void solve2() {
	if (n == 2) {
		cout << -1 << endl; return;
	}
	vector<int> a(m), b(m);
	rep(i, m) {
		cin >> a[i] >> b[i]; a[i]--; b[i]--;
	}
	string ans; ans.resize(m);
	vector<int> ids;
	vector<int> t;
	rep(i, m) {
		if (b[i] > 2) {
			ans[i] = '^';
		}
		else {
			t.push_back(a[i] + b[i] - 1);
			ids.push_back(i);
		}
	}
	rep(i, ids.size()) {
		if (i + 1 < ids.size()) {
			ans[ids[i]] = s3[t[i]][t[i + 1]];
		}
		else {
			ans[ids[i]] = 'v';
		}
	}
	cout << ans << endl;
}

void solve() {
	cin >> n >> m >> t;
	if (t == 2)solve2();
	else solve1();
}


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(12);
	//init();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}