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:
// ※※※ 解答不能 ※※※
// ksun48氏
// https://atcoder.jp/contests/agc041/submissions/9192801
#include <bits/stdc++.h>
using namespace std;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int T;
    scanf("%d", &T);
    vector<int> x(m), y(m);
    rep(i, m){
        scanf("%d %d", &x[i], &y[i]);
        x[i]--; y[i]--;
    }
    if(T == 1){
        using B = bitset<50000>;
        vector<B> cur(n);
        rep(i, n) cur[i][i].flip();
        rep(i, m){
            cur[x[i]] |= cur[y[i]];
            cur[y[i]] |= cur[x[i]];
        }
        rep(i, n){
            if((int)cur[i].count() == n){
                vector<int> where(n, 0);
                where[i] = 1;
                string st(m, '.');
                repr(j, m - 1, 0){
                    st[j] = "^v"[where[x[j]] < where[y[j]]];
                    where[x[j]] |= where[y[j]];
                    where[y[j]] |= where[x[j]];
                }
                printf("%s\n", st.c_str());
                return 0;
            }
        }
        puts("-1");
    }else if(T == 2){
        if(n == 2){
            puts("-1");
            return 0;
        }
        string st(m, '^');
        vector<int> reach(3, 1);
        int last = -1;
        rep(i, m){
            if(y[i] >= 3) continue;
            int other = 3 - x[i] - y[i];
            if(!reach[other]){
                assert(last >= 0);
                st[last] ^= ('^' ^ 'v');
                reach[other] = true;
            }
            st[i] = '^';
            reach[x[i]] = true;
            reach[y[i]] = false;
            last = i;
        }
        printf("%s\n", st.c_str());
    }
    return 0;
}