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;
int t,n,m;
int x[200000];
int y[200000];
long long ost[200000];
int kol[200000];
long long md[60000][1000];
bool vis[200000];
char s[200000];
long long c=62;
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) {
        int l=0;
        int tl=-1;
        int r=n/c+1;
        for(int i=0;i<n;i++) {
            ost[i]=(1ll<<(i%c));
            kol[i]=i/c;
            md[i][kol[i]]=ost[i];
        }
        for(int i=0;i<m;i++) {
            int tot=0;
            for(int j=0;j<=r;j++) {
                md[x[i]][j]|=md[y[i]][j];
                md[y[i]][j]=md[x[i]][j];
                tot+=__builtin_popcountll(md[x[i]][j]);
            }
            if(tot==n) {
                tl=i;
                break;
            }
        }
        s[m]=0;
        if(tl==-1) {
            printf("-1");
            return 0;
        }
        for(int i=0;i<n;i++) vis[i]=false;
        for(int i=m-1;i>=0;i--) {
            if(i>tl) s[i]='^';
            else {
                if(i==tl) {
                    vis[x[i]]=vis[y[i]]=true;
                    s[i]='^';
                }
                else {
                    if(vis[x[i]]) {
                        s[i]='^';
                        vis[y[i]]=true;
                    }
                    else {
                        if(vis[y[i]]) {
                            s[i]='v';
                            vis[x[i]]=true;
                        }
                        else s[i]='^';
                    }
                }
            }
        }
        printf("%s",s);
        return 0;
    }
    else {
        if(n==2) {
            printf("-1");
            return 0;
        }
        int a=0,b1=1,b2=2;
        s[m]=0;
        int pi=-1;
        for(int i=0;i<m;i++) {
            if(y[i]>2) {
                s[i]='^';
            }
            else {
                if(pi==-1) {
                    s[i]='^';
                    b1=x[i];
                    b2=y[i];
                    pi=i;
                    a=3-b1-b2;
                }
                else {
                    if((a==x[i] && b1==y[i]) || (a==y[i] && b1==x[i])) {
                        if(b1<b2) {
                            s[pi]='v';
                            swap(b1,b2);
                        }
                        else {
                            s[pi]='^';
                            swap(b1,b2);
                        }
                    }
                    if(a==x[i] || a==y[i]) {
                        if(a==x[i]) s[i]='^';
                        else s[i]='v';
                        swap(a,b1);
                        pi=i;
                    }
                    else {
                        if(b1<b2) {
                            s[i]='^';
                            pi=i;
                        }
                        else {
                            s[i]='v';
                            pi=i;
                        }
                    }
                }
            }
        }
        printf("%s",s);
        return 0;
    }
    return 0;
}