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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <cstdint>
using namespace std;
typedef long long ll;
#define MP make_pair
#define PB push_back
#define inf 1000000007
#define mod 1000000007
#define rep(i,n) for(int i = 0; i < (int)(n); ++i)

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m,t;
    cin >> n >> m >> t;
    vector<pair<int,int> > p;
    rep(i,m){
        int a,b;
        cin >> a >> b;
        p.push_back(MP(a,b));
    }
    vector<char> res(m);
    map<pair<int,int>,int> mp;
    if(t==2){
        if(n==2){
            cout << -1 << endl;
            return 0;
        }
        vector<pair<int,int> > q;
        rep(i,m){
            if(p[i]==MP(1,2)){
                q.push_back(MP(0,i));
            }else if(p[i]==MP(1,3)){
                q.push_back(MP(2,i));
            }else if(p[i]==MP(2,3)){
                q.push_back(MP(1,i));
            }else{
                res[i] = '^';
            }
        }
        int pid = 0;
        int id = 0;
        int K = q.size();
        int pre = -1;
        char c; 
        while(id<K){
            if(pre==-1){
                pre = q[id].first;
            }else{
                int k = q[id].first;
                if(pre!=k){
                    if(pre==0&&k==1){
                        c = '^';
                    }
                    if(pre==0&&k==2){
                        c = 'v';
                    }
                    if(pre==1&&k==0){
                        c = 'v';
                    }
                    if(pre==1&&k==2){
                        c = '^';
                    }
                    if(pre==2&&k==0){
                        c = 'v';
                    }
                    if(pre==2&&k==1){
                        c = '^';
                    }
                    for(int i=pid;i<id;i++){
                        //cerr << q[i].second << " " << c << endl;
                        res[q[i].second] = c;
                    }
                    pid = id;
                    pre = k;
                }
            }
            id++;
        }
        for(int i=pid;i<K;i++){
            res[q[i].second] = '^';
        }
        rep(i,m){
            cout << res[i];
        }
        cout << endl;
    }else{
        vector<bitset<50000> > bit(n);
        rep(i,n){
            bit[i].set(i);
        }
        rep(i,m){
            int a = p[i].first;
            int b = p[i].second;
            a--;b--;
            bit[a] |= bit[b];
            bit[b] |= bit[a];
        }
        int goal =-1;
        rep(i,n){
            if(bit[i].count()==n){
                goal = i;
                break;
            }
        }
        if(goal==-1){
            cout << -1 << endl;
        }else{
            bitset<500000> res;
            res.set(goal);
            string ans;
            for(int i=m-1;i>=0;i--){
                int a = p[i].first;
                int b = p[i].second;
                a--;b--;
                if(res[a]){
                    ans.push_back('^');
                    res.set(b);
                }else if(res[b]){
                    res.set(a);
                    ans.push_back('v');
                }else{
                    ans.push_back('^');
                }
            }
            reverse(ans.begin(),ans.end());
            cout << ans << endl;
        }
    }
    return 0;
}