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>
#define fi first
#define se second
#define mp make_pair
#define ss(x) (int) x.size()
#define pb push_back
#define ll long long
#define cat(x) cerr << #x << " = " << x << endl
#define FOR(i, n) for(int i = 0; i < n; ++i)

using namespace std;	

const int nax = 100005;

int n, m, t;
int x[nax];
int y[nax];
int dif;
int ile[nax];
char s[nax];
int g[nax];

bitset <50005> bit[50005];
bitset <50005> c;

void go(int k) {
	c.reset();
	c[k] = 1;
	for(int i = m; 1 <= i; --i) {
		if(c[x[i]] == c[y[i]])
			s[i] = '^';
		else if(c[x[i]] < c[y[i]]) {
			s[i] = 'v';
			c[x[i]] = 1;
		}
		else if(c[x[i]] > c[y[i]]) {
			s[i] = '^';
			c[y[i]] = 1;
		}
	}
	for(int i = 1; i <= m; ++i)
		printf("%c", s[i]);
	exit(0);
}
		
int T[nax];	

void solve1() {
	for(int i = 1; i <= n; ++i)
		bit[i][i] = 1;
	for(int i = m; 1 <= i; --i) {
		c = (bit[x[i]] | bit[y[i]]);
		bit[x[i]] = c;
		bit[y[i]] = c;
	}
	
	c.set();
	for(int i = 1; i <= n; ++i) 
		c = (c & bit[i]);
	for(int i = 1; i <= n; ++i)
		if(c[i] == 1)
			go(i);
			
	printf("-1\n");
}

void solve2() {
	if(n == 2) {
		printf("-1\n");
		return;
	}
	for(int i = 1; i <= n; ++i) {
		ile[i] = 1;
		g[i] = i;
	}
	dif = n;
	for(int i = m; 1 <= i; --i) {
		int xp = g[x[i]];
		int yp = g[y[i]];
		if(ile[xp] < ile[yp]) {
			ile[yp]--;
			ile[xp]++;
			g[y[i]] = xp;
			s[i] = '^';
		}
		else {
			ile[yp]++;
			ile[xp]--;
			g[x[i]] = yp;
			s[i] = 'v';
		}
	}
	for(int i = 1; i <= m; ++i)
		printf("%c", s[i]);
		
}
			
int main() {
	scanf("%d %d %d", &n, &m, &t);
	for(int i = 1; i <= m; ++i) {
		scanf("%d %d", x + i, y + i);
	}
	if(t == 1)
		solve1();
	else
		solve2();
	
	return 0;
}