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	LL	long long
#define	INF	0x3f3f3f3f
using namespace std;

template <typename T> void chkmax(T &x, T y){x = max(x, y);}
template <typename T> void chkmin(T &x, T y){x = min(x, y);}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
const int N = 50050,M = 100050;
int n,m,T;
int ex[M],ey[M]; char ans[M]; bitset<50001>v[N];
int main(){
	int i,j; 
	bool vis[N];
	read(n),read(m),read(T);
	for (i = 1; i <= m; ++i) read(ex[i]),read(ey[i]);
	if (T==2){
		int cnt[N],a[N];
		if (n==2){puts("-1");return 0;}
		for (i = 1; i <= n; ++i) cnt[i] = 1,a[i] = i;
		for (i = m; i >= 1; --i){
			int x = ex[i],y = ey[i];
			if (a[x]==a[y]) ans[i]='^';
			else if (cnt[a[x]] >= 2){ --cnt[a[x]]; a[x] = a[y]; ++cnt[a[x]]; ans[i]='v'; }
			else{ --cnt[a[y]]; a[y] = a[x]; ++cnt[a[y]]; ans[i]='^'; }
		}
		for (i = 1; i <= m; ++i) putchar(ans[i]); putchar('\n');
		return 0;
	}
	for (i = 1; i <= n; ++i) v[i].set(i);
	for (i = 1; i <= m; ++i) v[ex[i]] = v[ey[i]] = v[ex[i]] | v[ey[i]];
	for (i = 1; i <= n; ++i) if (v[i].count()==n){
		vis[i] = 1;
		for (j = m; j >= 1; --j) if (vis[ex[j]]) ans[j] = '^',vis[ey[j]] = 1; else if (vis[ey[j]]) ans[j] = 'v',vis[ex[j]] = 1; else ans[j] = 'v';
		for (j = 1; j <= m; ++j) putchar(ans[j]); putchar('\n');
		return 0;
	}
	puts("-1");
	return 0;
}