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;

const int MAXN = int(5e4) + 2;
const int MAXM = int(1.1e5);
int N, M;
int T;
pair<int, int> X[MAXM];

bool ansUp[MAXM];

using bs = bitset<MAXN>;

namespace uniforming {
bs canReach[MAXN];
bool vis[MAXN];
bool findUniforming() {
	for (int t = 0; t < N; t++) canReach[t][t] = true;

	for (int i = M-1; i >= 0; i--) {
		int x = X[i].first, y = X[i].second;
		canReach[x] |= canReach[y];
		canReach[y] = canReach[x];
	}

	bs isGood;
	for (int t = 0; t < N; t++) {
		isGood[t] = true;
	}
	for (int s = 0; s < N; s++) {
		isGood &= canReach[s];
	}

	for (int t = 0; t < N; t++) {
		if (isGood[t]) {
			vis[t] = true;
			for (int i = M-1; i >= 0; i--) {
				int x = X[i].first, y = X[i].second;
				if (vis[x] || vis[y]) {
					if (!vis[y]) {
						ansUp[i] = true;
						vis[y] = true;
					}
					if (!vis[x]) {
						ansUp[i] = false;
						vis[x] = true;
					}
					assert(vis[x] && vis[y]);
				}
			}
			for (int s = 0; s < N; s++) assert(vis[s]);
			return true;
		}
	}
	return false;
}
}

namespace nonUniforming {
bool findNonUniforming() {
	if (N == 2 && M >= 1) {
		return false;
	}
	int badState = 0;
	for (int i = M-1; i >= 0; i--) {
		int x = X[i].first, y = X[i].second;
		if (y >= 3) {
			ansUp[i] = true;
			continue;
		}
		if (x == 0 && y == 1) {
			if (badState == 1) ansUp[i] = false;
			else if (badState == 2) ansUp[i] = true;
			badState = 0;
		} else if (x == 0 && y == 2) {
			if (badState == 0) ansUp[i] = false;
			else if (badState == 2) ansUp[i] = true;
			badState = 1;
		} else if (x == 1 && y == 2) {
			if (badState == 0) ansUp[i] = false;
			else if (badState == 1) ansUp[i] = true;
			badState = 2;
		} else assert(false);
	}
	return true;
}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> M >> T;
	for (int i = 0; i < M; i++) {
		cin >> X[i].first >> X[i].second;
		X[i].first--, X[i].second--;
	}

	bool found;
	if (T == 1) {
		found = uniforming::findUniforming();
	} else if (T == 2) {
		found = nonUniforming::findNonUniforming();
	} else assert(false);
	if (found) {
		for (int i = 0; i < M; i++) {
			cout << "v^"[ansUp[i]];
		}
		cout << '\n';
	} else {
		cout << -1 << '\n';
	}

	return 0;
}