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>
#define Maxn 50007
#define Maxm 100007
using namespace std;
int n,m,T;
int x[Maxm],y[Maxm];
bitset<Maxn> f[Maxn];
vector<int> q[Maxn];
int id[Maxn];
bool vis[Maxn],ans[Maxm];
void solve1()
{
    for (int i=1;i<=n;i++)
    {
        f[i]=0;
        f[i][i]=1;
    }
    for (int i=1;i<=m;i++)
    {
        f[x[i]]=f[x[i]]|f[y[i]];
        f[y[i]]=f[x[i]];
    }
    int pos=-1;
    for (int i=1;i<=n;i++)
        if (f[i].count()==n)
        {
            pos=i;
            break;
        }
    if (pos==-1)
    {
        printf("%d\n",-1);
        return;
    }
    memset(vis,false,sizeof(vis));
    memset(ans,false,sizeof(ans));
    vis[pos]=true;
    for (int i=m;i;i--)
    {
        if (vis[x[i]]&&(!vis[y[i]])) ans[i]=true;
        if (vis[x[i]]||vis[y[i]])
        {
            vis[x[i]]=true;
            vis[y[i]]=true;
        }
    }
    for (int i=1;i<=m;i++)
        if (ans[i]) printf("^"); else printf("v");
    printf("\n");
}
void solve2()
{
    if (n==2)
    {
        printf("%d\n",-1);
        return;
    }
    memset(q,0,sizeof(q));
    for (int i=1;i<=m;i++)
    {
        q[x[i]].push_back(y[i]);
        q[y[i]].push_back(x[i]);
    }
    memset(id,0,sizeof(id));
    int a=x[1],b=y[1];
    b=b%n+1;
    if (a==b) b=b%n+1;
    for (int i=1;i<=m;i++)
    {
        ++id[x[i]];
        ++id[y[i]];
        if (x[i]!=a&&x[i]!=b&&y[i]!=a&&y[i]!=b) printf("^");
        else 
        {
            if (x[i]==b||y[i]==b) swap(a,b);
            if (id[b]==(int)q[b].size()||q[b][id[b]]!=x[i]+y[i]-a)
            {
                if (x[i]==a) printf("v"); else printf("^");
                a=x[i]+y[i]-a;
            } else
            {
                if (x[i]==a) printf("^"); else printf("v");
            }
        }
    }
    printf("\n");
}
int main()
{
    scanf("%d%d%d",&n,&m,&T);
    for (int i=1;i<=m;i++)
        scanf("%d%d",&x[i],&y[i]);
    if (T==1) solve1(); else solve2();
    return 0;
}