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>

const int N = 50010;
const int M = 100010;

std::bitset <N> set[N];
int n, m, t;
int x[M], y[M];
bool vis[N];
char ans[M];
int cnt[N], target[N];

void solve1(){
    for (int i = 0; i < n; ++ i){
        set[i][i] = true;
    }
    for (int i = m - 1; i >= 0; -- i){
        set[x[i]] = set[y[i]] = set[x[i]] | set[y[i]];
    }
    std::bitset <N> set1;
    set1.set();
    for (int i = 0; i < n; ++ i){
        set1 &= set[i];
    }
    for (int i = 0; i < n; ++ i){
        if (set1[i]){
            vis[i] = true;
            for (int j = m - 1; j >= 0; -- j){
                if (int(vis[x[j]]) + int(vis[y[j]]) != 1){
                    ans[j] = '^';
                    continue;
                }
                if (!vis[x[j]]){
                    ans[j] = 'v';
                }
                else{
                    ans[j] = '^';
                }
                vis[x[j]] = vis[y[j]] = vis[x[j]] || vis[y[j]];
            }
            puts(ans);
            return;
        }
    }
    puts("-1");
}

void solve2(){
    if (n == 2){
        puts("-1");
        return;
    }
    for (int i = 0; i < n; ++ i){
        cnt[i] = 1;
        target[i] = i;
    }
    for (int i = m - 1; i >= 0; -- i){
        int &t1 = target[x[i]], &t2 = target[y[i]];
        if (cnt[t1] + cnt[t2] < n || cnt[t2] >= 2){
            ans[i] = '^';
            -- cnt[t2];
            ++ cnt[t1];
            t2 = t1;
        }
        else{
            ans[i] = 'v';
            -- cnt[t1];
            ++ cnt[t2];
            t1 = t2;
        }
    }
    puts(ans);
}

int main(){
    scanf("%d%d%d", &n, &m, &t);
    for (int i = 0; i < m; ++ i){
        scanf("%d%d", &x[i], &y[i]);
        -- x[i], -- y[i];
    }
    if (t == 1){
        solve1();
    }
    else{
        solve2();
    }
    return 0;
}