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; 
typedef double db; 
typedef string str; 

typedef pair<int, int> pi;
typedef pair<ll,ll> pl; 
typedef pair<ld,ld> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl;
typedef vector<ld> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x) 
#define rsz resize
#define ins insert 

#define mp make_pair
#define pb push_back
#define eb emplace_back 
#define f first
#define s second
#define lb lower_bound 
#define ub upper_bound 

const int MOD = 1e9+7; // 998244353; // = (119<<23)+1
const int MX = 2e5+5;
const ll INF = 1e18; 
const ld PI = 4*atan((ld)1); 

template<class T> bool ckmin(T& a, const T& b) { 
	return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; }

mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

namespace input {
	template<class T> void re(complex<T>& x);
	template<class T1, class T2> void re(pair<T1,T2>& p);
	template<class T> void re(vector<T>& a);
	template<class T, size_t SZ> void re(array<T,SZ>& a);

	template<class T> void re(T& x) { cin >> x; }
	void re(double& x) { string t; re(t); x = stod(t); }
	void re(ld& x) { string t; re(t); x = stold(t); }
	template<class T, class... Ts> void re(T& t, Ts&... ts) { 
		re(t); re(ts...); 
	}

	template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
	template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
	template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
	template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
	void pr(int x) { cout << x; }
	void pr(long x) { cout << x; }
	void pr(ll x) { cout << x; }
	void pr(unsigned x) { cout << x; }
	void pr(unsigned long x) { cout << x; }
	void pr(unsigned long long x) { cout << x; }
	void pr(float x) { cout << x; }
	void pr(double x) { cout << x; }
	void pr(ld x) { cout << x; }
	void pr(char x) { cout << x; }
	void pr(const char* x) { cout << x; }
	void pr(const string& x) { cout << x; }
	void pr(bool x) { pr(x ? "true" : "false"); }
	template<class T> void pr(const complex<T>& x) { cout << x; }
	
	template<class T1, class T2> void pr(const pair<T1,T2>& x);
	template<class T> void pr(const T& x);
	
	template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { 
		pr(t); pr(ts...); 
	}
	template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
		pr("{",x.f,", ",x.s,"}"); 
	}
	template<class T> void pr(const T& x) { 
		pr("{"); // const iterator needed for vector<bool>
		bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
		pr("}");
	}
	
	void ps() { pr("\n"); } // print w/ spaces
	template<class T, class... Ts> void ps(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(" "); ps(ts...); 
	}
	
	void pc() { pr("]\n"); } // debug w/ commas
	template<class T, class... Ts> void pc(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(", "); pc(ts...); 
	}
	#define dbg(x...) pr("[",#x,"] = ["), pc(x);
}

using namespace output;

namespace io {
	void setIn(string s) { freopen(s.c_str(),"r",stdin); }
	void setOut(string s) { freopen(s.c_str(),"w",stdout); }
	void setIO(string s = "") {
		ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
		// cin.exceptions(cin.failbit); // ex. throws exception when you try to read letter into int
		if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
	}
}

using namespace io;

bitset<50001> cur[50001], ok;
int N,M,T;
vpi v;

void fin(int x) {
	ok[x] = 1;
	str res(M,'.');
	R0F(i,sz(v)) {
		auto a = v[i];
		if (ok[a.f]) {
			res[i] = '^';
			if (ok[a.f]) ok[a.s] = 1;
		} else {
			res[i] = 'v';
			if (ok[a.s]) ok[a.f] = 1;
		}
	}
	FOR(i,1,N+1) assert(ok[i]);
	ps(res);
	exit(0);
}

int fin(vi a, int b) {
	int ind = 0; while (ind < sz(a) && a[ind] != b) ind ++;
	if (ind == sz(a)) return -1;
	return ind;
}

bool con(vi a, int b) {
	return fin(a,b) != -1;
}

void rem(vi& a, int b) {
	int ind = fin(a,b); assert(ind != -1);
	a.erase(begin(a)+ind);
}

void tow(char& c, pi v, int& p) {
	if (v.f == p) c = '^';
	else if (v.s == p) c = 'v';
}

void away(char& c, pi v, int& p) {
	if (v.f == p) c = '^', p = v.s;
	else if (v.s == p) c = 'v', p = v.f;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	re(N,M,T);
	if (T == 1) {
		FOR(i,1,N+1) cur[i][i] = 1;
		F0R(i,M) {
			int x,y; re(x,y); v.pb({x,y});
			cur[x] = cur[y] = cur[x]|cur[y];
		}
		FOR(i,1,N+1) if (cur[i].count() == N) fin(i);
	} else {
		if (N == 2) {
			ps(-1);
			exit(0);
		}
		F0R(i,M) {
			int x,y; re(x,y); v.pb({x,y});
		}
		vector<vi> a, b;
		a.pb({v[0].f,v[0].s});
		b.eb();
		FOR(i,1,N+1) if (i != v[0].f && i != v[0].s) {
			b.back().pb(i);
			break;
		}
		FOR(i,1,sz(v)) {
			a.pb(a.back()), b.pb(b.back());
			auto t = v[i];
			if (!con(a.back(),t.f) || !con(b.back(),t.s)) swap(t.f,t.s);
			if (!con(a.back(),t.f) || !con(b.back(),t.s)) continue;
			if (sz(a.back()) == 2) {
				rem(a.back(),t.f);
				b.back() = {t.f,t.s};
			} else {
				assert(sz(b.back()) == 2);
				rem(b.back(),t.s);
				a.back() = {t.f,t.s};
			}
		}
		str res(M,'^');
		// ps("HA",a,b);
		// exit(0);
		int pa = a.back()[0], pb = b.back()[0];
		ROF(i,1,M) {
			//ps("WHOOPS",i,pa,pb,v[i]);
			if (con(a[i-1],pa) && con(b[i-1],pb)) {
				tow(res[i],v[i],pa);
				tow(res[i],v[i],pb);
			} else {
				//ps("START",pa,pb);
				if (!con(a[i-1],pa))  {
					//ps("OPT1");
					away(res[i],v[i],pa);
				} else {
					//ps("OPT2");
					away(res[i],v[i],pb);
				}
				if (!(con(a[i-1],pa) && con(b[i-1],pb))) {
					//ps("OOPS",a[i],b[i]);
					ps(a[i-1],b[i-1]);
					ps(pa,pb);
					ps(res[i],v[i]);
					exit(5);
				}
			}
			// if (con(a[i-1],))
		}
		tow(res[0],v[0],pa);
		assert(pa != pb);
		// ps(pa,pb);
		ps(res);
		exit(0);
	}
	ps(-1);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?), slow multiset operations
	* do smth instead of nothing and stay organized
*/