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;
using ll = long long;

void ch(int &x, int &y){
    if(y < x){
        int tmp = y;
        y = x;
        x = tmp;
    }
}

int N, M, T;
vector<int>x, y;

void solve1(){
    using bs = bitset<50001>;
    vector<bs>dp(N+1); // dp[i][j] は i から j にゴールできるかどうか
    for (int i = 1; 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]];
    }

    bs flag1; flag1.set();
    for (int i = 1; i <= N; ++i) {
        flag1 &= dp[i];
    }

    for (int j = 1; j <= N; ++j) {
        if (flag1[j]) {
            string s(M, '^');
            bs flag;
            flag[j] = 1;
            for (int i = M - 1; i >= 0; --i) {
                if (flag[y[i]] == 1) s[i] = 'v';
                flag[x[i]] = flag[x[i]] || flag[y[i]];
                flag[y[i]] = flag[x[i]];
            }
            assert(flag.count() == N);
            cout << s << "\n";
            return;
        }
    }
    cout << -1 << "\n";
}

void solve2(){
    if(N == 2){
        cout << -1 << "\n";
        return;
    }
    vector<vector<int>>graph(N+1, vector<int>());
    for (int i = 0; i < M; ++i) {
        graph[x[i]].push_back(i);
        graph[y[i]].push_back(i);
    }

    int x1 = x[0];
    int y1 = y[0] + 1;
    if(y1 == N + 1) {
        if(x1 == 1) y1 = 2;
        else y1 = 1;
    }
    ch(x1, y1);

    // 次で衝突する場合は進路転換
    for (int i = 0; i < M; ++i) {
        auto iter1 = upper_bound(graph[x1].begin(), graph[x1].end(), i);
        auto iter2 = upper_bound(graph[y1].begin(), graph[y1].end(), i);
        if(iter1 != graph[x1].end() && iter2 != graph[y1].end() && (*iter1 == *iter2)){
            if(x[i] == x1){
                cout << "v";
                x1 = y[i];
            } else if(y[i] == y1){
                cout << "^";
                y1 = x[i];
            } else if(y[i] == x1){
                cout << "^";
                x1 = x[i];
            } else if(x[i] == y1){
                cout << "v";
                y1 = y[i];
            }
        } else {
            if (y[i] == y1 || y[i] == x1) {
                cout << "v";
            } else {
                cout << "^";
            }
        }
        ch(x1, y1);
    }
}

int main() {
    cin >> N >> M >> T;
    x.resize(M); y.resize(M);
    for (int i = 0; i < M; ++i) {
        cin >> x[i] >> y[i];
    }
    if (T == 1) solve1();
    else solve2();
    return 0;
}