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;
bitset<50005> st[50005];
const int Q=1<<20;
int X[Q],Y[Q];
int had[Q];
char ans[Q];
int si[Q],bel[Q];
int main()
{
    int n,m,ty;
    scanf("%d%d%d",&n,&m,&ty);
    for(int i=1;i<=m;i++)
            scanf("%d%d",&X[i],&Y[i]);
    if(ty==1){
        for(int i=1;i<=n;i++)
            st[i][i]=1;
        for(int i=1;i<=m;i++)
            st[X[i]]=st[Y[i]]=st[X[i]]|st[Y[i]];
        int beg=-1;
        for(int i=1;i<=n;i++)
            if(st[i].count()==n){
                beg=i;
                break;
            }
        if(beg<0)return 0&puts("-1");
        had[beg]=1;
        for(int i=m;i;--i)
            if(!had[X[i]])ans[i]='v',had[X[i]]|=had[Y[i]];
            else ans[i]='^',had[Y[i]]|=had[X[i]];
        ans[m+1]='\n';
        printf("%s",ans+1);
        return 0;
    }
    if(n==2)return 0&puts("-1");
    for(int i=1;i<=n;i++)
        bel[i]=i,si[i]=1;
    for(int j=m;j;--j){
        int tx=bel[X[j]],ty=bel[Y[j]];
        if(tx==ty)ans[j]='^';
        else if(si[tx]>=2)ans[j]='v',--si[tx],++si[ty],bel[X[j]]=ty;
        else ans[j]='^',++si[tx],--si[ty],bel[Y[j]]=tx;
    }
    ans[m+1]='\n';
    printf("%s",ans+1);
    return 0;
}