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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<bitset>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;


vector<Int> edge[55000];
bitset<50000> bits[55000];
Int x[110000], y[110000];
int main(){
    Int n, m, t, t1 = -1, t2 = -1;
    cin >> n >> m >> t;
    for(int i = 0;i < m;i++){
        cin >> x[i] >> y[i];x[i]--,y[i]--;
        edge[x[i]].push_back(y[i]);
        edge[y[i]].push_back(x[i]);
    }
    if(t == 1){
        for(int i = 0;i < n;i++){
            bits[i].set(i);
        }
        for(int i = m-1;i >= 0;i--){
            bits[x[i]] |= bits[y[i]];
            bits[y[i]] |= bits[x[i]];
        }
        int t = -1;
        for(int i = 0;i < n;i++){
            bits[0] &= bits[i]; 
        }
        for(int i = 0;i < n;i++){
            if(bits[0][i]){
                t = i;
            }
        }
        if(t == -1){
            cout << -1 << endl;
            return 0;
        }
        vector<int> ok(n, 0);
        string ans(m, 'v');
        ok[t] = 1;
        for(int i = m-1;i >= 0;i--){
            if(ok[x[i]])ans[i] = '^';
            ok[x[i]] |= ok[y[i]];
            ok[y[i]] |= ok[x[i]];
        }
        cout << ans << endl;
        return 0;
    }

    if(n == 2){
        cout << -1 << endl;
        return 0;
    }
    vector<int> to(n, 0), cnt(n, 1);
    string ans(m, '^');
    for(int i = 0;i < n;i++){
        to[i]  = i;
    }
    for(int i = m-1;i >= 0;i--){
        if(cnt[to[x[i]]] > 1){
            ans[i] = 'v';
            cnt[to[y[i]]]++;
            cnt[to[x[i]]]--;
            to[x[i]] = to[y[i]];
        }
        else{
            cnt[to[x[i]]]++;
            cnt[to[y[i]]]--;
            to[y[i]] = to[x[i]];
        }     //   for(auto elem:to)cout << elem << " ";cout << endl;
    }
    cout << ans << endl;
        
    
    return 0;
}