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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef long double ld;
#define mp make_pair
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int B = 64;
const int M = 50010;
const int S = M / B + 1;
const int N = 100100;
int n, m;
int a[N][2];
char ans[N];
bool used[N];
ull b[M][S];

void solveUniform() {
	for (int i = 0; i < n; i++)
		b[i][i / B] ^= (ull)1 << (i % B);
	for (int i = m - 1; i >= 0; i--) {
		int v = a[i][0], u = a[i][1];
		for (int j = 0; j < S; j++) {
			b[v][j] |= b[u][j];
			b[u][j] = b[v][j];
		}
	}
	for (int i = 1; i < n; i++)
		for (int j = 0; j < S; j++)
			b[0][j] &= b[i][j];
	int V = -1;
	for (int i = 0; V == -1 && i < S; i++) {
		if (b[0][i] == 0) continue;
		int j = 0;
		while(((b[0][i] >> j) & 1) == 0) j++;
		V = i * B + j;
	}
	if (V == -1) {
		printf("-1\n");
		return;
	}
	for (int i = 0; i < n; i++)
		used[i] = 0;
	used[V] = 1;
	for (int i = m - 1; i >= 0; i--) {
		int v = a[i][0], u = a[i][1];
		if (used[v]) {
			ans[i] = '^';
			used[u] |= used[v];
		} else {
			ans[i] = 'v';
			used[v] |= used[u];
		}
	}
	ans[m] = '\0';
	printf("%s\n", ans);
}

void solveNonuniform() {
	if (n == 2) {
		printf("-1\n");
		return;
	}
	for (int i = 0; i < n; i++)
		used[i] = 1;
	int x = a[0][0], y = a[0][1];
	used[x] = used[y] = 0;
	int id = 0;
	for (int i = 1; i < m; i++) {
		if (x > y) swap(x, y);
		//eprintf("%d %d %d\n", id, x, y);
		if (x == a[i][0] && y == a[i][1]) {
			ans[id] = '^';
			id = i;
			continue;
		}
		if (a[i][0] == x || a[i][1] == x || a[i][0] == y || a[i][1] == y) {
			if (a[i][0] == y || a[i][1] == y) swap(x, y);
			int z = x ^ a[i][0] ^ a[i][1];
			if (!used[z]) {
				ans[i] = (x < z ? '^' : 'v');
				continue;
			}
			ans[id] = (x < y ? 'v' : '^');
			used[y] = 1;
			used[x] = used[z] = 0;
			y = z;
			id = i;
		} else {
			ans[i] = '^';
			used[a[i][0]] |= used[a[i][1]];
			used[a[i][1]] = 0;
		}
	}
	ans[id] = 'v';
	ans[m] = '\0';
	printf("%s\n", ans);
}

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	int t;
	scanf("%d%d%d", &n, &m, &t);
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < 2; j++) {
			scanf("%d", &a[i][j]);
			a[i][j]--;
		}
	}
	if (t == 1)
		solveUniform();
	else
		solveNonuniform();

	return 0;
}