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<iostream>
#include<bitset>
#define DIM 50005
#define f first
#define s second
using namespace std;
int n, m, i, t, x;
int c[DIM], num[DIM];
bitset<DIM> a[DIM], d, aux;
char sol[2 * DIM];
pair<int, int> v[2 * DIM];
int main(){
    cin>> n >> m >> t;
    for(i = 1; i <= m; i++){
        cin>> v[i].f >> v[i].s;
    }
    if(t == 2){
        if(n == 2){
            cout<< -1;
            return 0;
        }
        for(i = 1; i <= n; i++){
            c[i] = i;
            num[i] = 1;
        }
        for(i = m; i >= 1; i--){
            if(num[ c[ v[i].f ] ] > num[ c[ v[i].s ] ]){
                sol[i] = 'v';
                num[ c[ v[i].f ] ]--;
                num[ c[ v[i].s ] ]++;
                c[ v[i].f ] = c[ v[i].s ];
            }
            else{
                sol[i] = '^';
                num[ c[ v[i].f ] ]++;
                num[ c[ v[i].s ] ]--;
                c[ v[i].s ] = c[ v[i].f ];
            }
        }
    }
    else{
        for(i = 1; i <= n; i++){
            a[i][i] = d[i] = 1;
        }
        for(i = m; i >= 1; i--){
            aux = (a[ v[i].f ] | a[ v[i].s ]);
            a[ v[i].f ] = a[ v[i].s ] = aux;
        }
        for(i = 1; i <= n; i++){
            d &= a[i];
        }
        for(i = 1; i <= n; i++){
            if(d[i] == 1){
                x = i;
                break;
            }
        }
        if(i == n + 1){
            cout<< -1;
            return 0;
        }
        c[x] = 1;
        for(i = m; i >= 1; i--){
            if(c[ v[i].f ] == 1){
                sol[i] = '^';
                c[ v[i].s ] = 1;
            }
            else{
                if(c[ v[i].s ] == 1){
                    sol[i] = 'v';
                    c[ v[i].f ] = 1;
                }
                else{
                    sol[i] = '^';
                }
            }
        }
    }
    cout<< sol + 1;
}