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 vector<int> VI;
typedef pair <int,int> ii;
typedef long long LL;
#define pb push_back
const int INF = 2147483647;
const int N = 50005;
const int M = 100005;

int n, m, t, i, x[M], y[M], mapa[N], co[N];
bitset<N> b[N], bb;
string s;
bool cc[N];


void solve() {
	if (n == 2) {
		printf("-1\n");
		return;
	}
	for (i=0;i<n;i++) {
		mapa[i] = i;
		co[i] = 1;
	}
	for (i=m-1;i>=0;i--) {
		if (co[mapa[x[i]]] > 1) {
			s += 'v';
			co[mapa[x[i]]]--;
			mapa[x[i]] = mapa[y[i]];
			co[mapa[x[i]]]++;
		} else {
			s += '^';
			co[mapa[y[i]]]--;
			mapa[y[i]] = mapa[x[i]];
			co[mapa[y[i]]]++;
		}
	}
	reverse(s.begin(), s.end());
	printf("%s\n", s.c_str());
}

int main() {
scanf("%d %d %d", &n, &m, &t);
s="";
for (i=0;i<m;i++) {
	scanf("%d %d", &x[i], &y[i]);
	x[i]--;
	y[i]--;
}
if (t == 2) {
	solve();
	return 0;
}
for (i=0;i<n;i++) {
	b[i].set(i);
	bb.set(i);
}
for (i=m-1;i>=0;i--) b[x[i]] = b[y[i]] = b[x[i]] | b[y[i]];
for (i=0;i<n;i++) bb &= b[i];
for (i=0;i<n;i++) if (bb[i]) {
	cc[i] = 1;
	for(i=m-1;i>=0;i--) {
		if (cc[y[i]]) s += 'v'; else s += '^';
		cc[x[i]] = cc[y[i]] = cc[x[i]] | cc[y[i]];
	}
	reverse(s.begin(), s.end());
	printf("%s\n", s.c_str());
	return 0;
}
printf("-1\n");
return 0;
}