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;
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<LL,LL>
#define x first
#define y second
#define pi acos(-1)
#define sqr(x) ((x)*(x))
#define pdd pair<double,double>
#define MEMS(x) memset(x,-1,sizeof(x))
#define MEM(x) memset(x,0,sizeof(x))
#define less Less
#define EPS 1e-4
#define arg ARG
#define cpdd const pdd
#define rank Rank
#define KK 500
#define MXN 200005
int f[50005];
int Find(int x){
    if(f[x]==x)return x;
    return f[x]=Find(f[x]);
}
bitset<50005> b[50005];
int main(){
    int n,m,t;
    scanf("%d %d %d",&n,&m,&t);
    if(t==2){
        if(n==2){
            printf("-1\n");
            return 0;
        }
        set<pii> s;
        pii p[100005];
        deque<pii> q;
        for(int i = 0;i<m;i++){
            int x,y;
            scanf("%d %d",&x,&y);
            //s.insert(mp(x,y));
            p[i]=mp(x,y);
            if(y<=3){
                if(q.empty()||q.back()!=mp(x,y))
                q.push_back(mp(x,y));
            }
        }
        char last;
        for(int i = 0;i<m;i++){
            if(p[i].y>3)printf("^");
            else{
                if(q.empty()||q.front()!=p[i]){
                    printf("%c",last);
                }
                else{
                    pii p=q.front();
                    q.pop_front();
                    if(q.empty())printf("^"),last='^';
                    else{
                        if(p.x==q.front().x||p.x==q.front().y){
                            printf("v");last='v';
                        }
                        else{
                            printf("^");last='^';
                        }
                    }
                }
            }
        }
        printf("\n");
    }
    else{
     /*   for(int i = 1;i<=n;i++){
            f[i]=i;
        }*/

        for(int i = 1;i<=n;i++)
            b[i][i]=1;
        pii p[100005];
        vector<int> q;
        int cnt=0;
        for(int i = 1;i<=m;i++){
            int x,y;
           // int x=rand()%n+1,y=rand()%n+1;
            scanf("%d %d",&x,&y);
        //    b[x]|=b[y];
         //   b[y]|=b[x];
         /*   while(y==x){
                y=rand()%n+1;
            }
            if(y<x)swap(x,y);*/
           /* printf("%d %d\n",x,y);
            p[i]=mp(x,y);
            int fx=Find(x);
            int fy=Find(y);
            if(fx!=fy){
                q.pb(i);
                f[fx]=fy;
                printf("!\n");
            }*/
            p[i]=mp(x,y);
            b[x]|=b[y];
            b[y]|=b[x];
        }
        for(int i = 1;i<=n;i++){
            if(b[i].count()==n){
              //  printf("%d\n",i);
                int ok[100005];
                MEM(ok);
                ok[i]=1;
                char c[100005];
                MEM(c);
                for(int j=m;j>=1;j--){
                  //  printf("%d %d\n",p[j].x,ok[p[j].x]);
                    if(ok[p[j].x]){
                        c[j]='^';
                        //printf("^");
                        ok[p[j].y]=1;
                    }
                    else{
                        if(ok[p[j].y]){
                            ok[p[j].x]=1;
                        }
                        c[j]='v';
                    }
                }
                printf("%s\n",c+1);
                return 0;
            }
        }
        printf("-1\n");
    }
}