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 sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
using lint = long long;
using pi = pair<int, int>;
const int MAXN = 100005;

int n, m;
pi a[MAXN];
int dp[MAXN][3][3];

void solve(){
	if(n == 2){
		puts("-1");
		return;
	}
	for(int i=0; i<3; i++){
		for(int j=0; j<i; j++){
			dp[m][i][j] = 1;
			dp[m][j][i] = 1;
		}
	}
	for(int i=m-1; i>=0; i--){
		if(a[i].second >= 3){
			memcpy(dp[i], dp[i + 1], sizeof(dp[i]));
			continue;
		}
		for(int j=0; j<3; j++){
			for(int k=0; k<j; k++){
				if(pi(k, j) == a[i]) continue;
				if(j == a[i].first || j == a[i].second){
					dp[i][j][k] |= dp[i + 1][j][k];
					dp[i][j][k] |= dp[i + 1][a[i].first + a[i].second - j][k];
				}
				if(k == a[i].first || k == a[i].second){
					dp[i][j][k] |= dp[i + 1][j][k];
					dp[i][j][k] |= dp[i + 1][a[i].first + a[i].second - k][j];
				}
				dp[i][k][j] = dp[i][j][k];
			}
		}
	}
	int px = -1, py = -1;
	for(int i=0; i<3; i++){
		for(int j=0; j<3; j++){
			if(dp[0][i][j]){
				px = i, py = j;
			}
		}
	}
	if(px == -1) assert(0);
	for(int i=0; i<m; i++){
		if(a[i].second >= 3){
			putchar('^');
			continue;
		}
		int j = px, k = py;
		if(j == a[i].first || j == a[i].second){
			bool val = (j == a[i].first) ^ (!dp[i + 1][j][k]);
			if(!val){
				putchar('v');
				if(j == a[i].first) j = a[i].second;
			}
			else{
				putchar('^');
				if(j == a[i].second) j = a[i].first;
			}
		}
		else if(k == a[i].first || k == a[i].second){
			bool val = (k == a[i].first) ^ (!dp[i + 1][j][k]);
			if(!val){
				putchar('v');
				if(k == a[i].first) k = a[i].second;
			}
			else{
				putchar('^');
				if(k == a[i].second) k = a[i].first;
			}
		}
		else putchar('^');
		px = j;
		py = k;
	}
	puts("");
}

bitset<50005> bs[50005];

void report(int x){
	bool chk[MAXN] = {};
	char ans[MAXN] = {};
	chk[x] = 1;
	for(int i=m-1; i>=0; i--){
		if(chk[a[i].first] == chk[a[i].second]){
			ans[i] = '^';
			continue;
		}
		if(chk[a[i].first]) ans[i] = '^';
		else ans[i] = 'v';
		chk[a[i].first] = chk[a[i].second] = 1;
	}
	cout << ans << endl;
}

int main(){
	int t;
	scanf("%d %d %d",&n,&m,&t);
	for(int i=0; i<m; i++){
		scanf("%d %d",&a[i].first,&a[i].second);
		a[i].first--;
		a[i].second--;
	}
	if(t == 1){
		for(int i=0; i<n; i++) bs[i].set(i);
		for(int i=0; i<m; i++){
			bs[a[i].first] = bs[a[i].second] = (bs[a[i].first] | bs[a[i].second]);
		}
		for(int i=0; i<n; i++){
			if(bs[i].count() == n){
				report(i);
				return 0;
			}
		}
		puts("-1");
		return 0;
	}
	if(t == 2){
		solve();
		return 0;
	}
}