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>
using namespace std;

int N, M, T;
int X[100000], Y[100000];

void solve1(){
    static bitset<50000> dp[50000], ok;
    for(int i=0; i<N; i++) dp[i][i] = 1;

    for(int i=M-1; i>=0; i--){
        dp[X[i]] |= dp[Y[i]];
        dp[Y[i]] = dp[X[i]];
    }
    ok.set();
    for(int i=0; i<N; i++) ok &= dp[i];

    for(int s=0; s<N; s++) if(ok[s]){
        bitset<50000> avail;
        string ans(M, '^');
        avail[s] = 1;
        for(int i=M-1; i>=0; i--){
            if(avail[X[i]]){
                avail[Y[i]] = 1;
            }else if(avail[Y[i]]){
                avail[X[i]] = 1;
                ans[i] = 'v';
            }
        }
        cout << ans << endl;
        return;
    }
    cout << -1 << endl;
}

void solve2(){
    if(N == 2){
        cout << -1 << endl;
        return;
    }
    vector<int> dest(N);
    vector<int> num(N, 1);
    for(int i=0; i<N; i++) dest[i] = i;
    string ans(M, ' ');
    for(int i=M-1; i>=0; i--){
        if(num[dest[X[i]]] >= 2){
            ans[i] = 'v';
            num[dest[X[i]]]--;
            num[dest[Y[i]]]++;
            dest[X[i]] = dest[Y[i]];
        }else{
            ans[i] = '^';
            num[dest[Y[i]]]--;
            num[dest[X[i]]]++;
            dest[Y[i]] = dest[X[i]];
        }
    }
    cout << ans << endl;
}

int main(){
    cin >> N >> M >> T;
    for(int i=0; i<M; i++){
        cin >> X[i] >> Y[i];
        X[i]--; Y[i]--;
    }
    if(T == 1) solve1();
    else solve2();
    
    return 0;
}