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 cs const
#define pb push_back
using namespace std;
cs int N = 5e4 + 50, M = 1e5 + 50;
int n, m, T, x[M], y[M];
bitset<N> S[N]; char ans[N];
void work(int t){
	static bool ok[N];
	ok[t] = true;
	for(int i=m; i>=1; i--)
	if(ok[x[i]]) ans[i] = '^', ok[y[i]] = true;
	else ans[i] = 'v', ok[x[i]] |= ok[y[i]];
	for(int i=1; i<=m; i++)
	putchar(ans[i]);
} int ct[N], tr[N];
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d%d%d",&n,&m,&T);
	for(int i=1; i<=m; i++)
	scanf("%d%d",&x[i],&y[i]);
	if(T==1){
		for(int i=1; i<=n; i++)
		S[i].set(i);
		for(int i=m; i>=1; i--)
		S[x[i]]=S[y[i]]=S[x[i]]|S[y[i]];
		bitset<N> t; t.set();
		for(int i=1; i<=n; i++)
		t = t & S[i]; 
		for(int i=1; i<=n; i++)
		if(t.test(i)) return work(i), 0;
		return puts("-1"), 0; 
	} else{
		if(n<=2) return puts("-1"), 0;
		for(int i=1; i<=n; i++) ct[i]=1, tr[i]=i;
		for(int i=m, o, e; i>=1; i--)
		if(ct[o=tr[x[i]]] > ct[e=tr[y[i]]])
		tr[x[i]] = tr[y[i]], ans[i] = 'v', --ct[o], ++ct[e];
		else ++ct[o], --ct[e], tr[y[i]] = tr[x[i]], ans[i] = '^';
		for(int i=1; i<=m; i++) putchar(ans[i]);
		return 0;
	}
}