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
#include<cstdio>
#include<algorithm>
#include<bitset>
using namespace std;
bitset<50000>B[50000];
int n, m, T, chk[50100];
struct Edge {
	int a, b;
}Ed[101000];
char res[100100];
int main() {
	int i;
	scanf("%d%d%d", &n, &m, &T);
	if (T == 1) {
		for (i = 0; i < n; i++)B[i][i] = 1;
		for (i = 0; i < m; i++) {
			int a, b;
			scanf("%d%d", &a, &b);
			a--, b--;
			Ed[i] = { a,b };
			B[a] = B[b] = (B[a] | B[b]);
		}
		int root = -1;
		for (i = 0; i < n; i++) {
			if (B[i].count() == n) {
				root = i;
				break;
			}
		}
		if (root == -1) {
			puts("-1");
			return 0;
		}
		chk[root] = 1;
		for (i = m - 1; i >= 0; i--) {
			if (chk[Ed[i].a]) {
				res[i] = '^';
			}
			else if (chk[Ed[i].b]) {
				res[i] = 'v';
			}
			else {
				res[i] = '^';
			}
			int t = chk[Ed[i].a] | chk[Ed[i].b];
			chk[Ed[i].a] = chk[Ed[i].b] = t;
		}
		printf("%s\n", res);
		return 0;
	}

	for (i = 0; i < m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a--, b--;
		Ed[i] = { a,b };
	}
	if (n == 2) {
		puts("-1");
		return 0;
	}
	int chk = 0;
	int b0, b1, p0, p1, last = -1, tb0, tb1, tp0, tp1;
	for (i = 0; i < m; i++) {
		if (Ed[i].b > 2) {
			res[i] = '^';
			continue;
		}
		int a = Ed[i].a, b = Ed[i].b;
		if (!chk) {
			chk = 1;
			b0 = 7 - (1 << Ed[i].b);
			p0 = 0;
			b1 = 7 - (1 << Ed[i].a);
			p1 = 1;
			last = i;
			continue;
		}
		int bb = (1 << a) + (1 << b);
		if (bb != b0) {
			tb0 = b0;
			tb1 = b0 ^ bb;
			if (!p0)res[last] = '^';
			else res[last] = 'v';
		}
		else {
			tb0 = b1;
			tb1 = b1 ^ bb;
			if (!p1)res[last] = '^';
			else res[last] = 'v';
		}
		if (tb0 < tb1) tp0 = 0, tp1 = 1;
		else tp0 = 1, tp1 = 0;
		b0 = tb0, p0 = tp0;
		b1 = tb1, p1 = tp1;
		last = i;
	}
	if (last != -1) {
		res[last] = '^';
	}
	printf("%s\n", res);
}