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;

typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back

const int MN = 5e4+5, MM = 1e5+5;
int T, N, M, i, x, y, st, ds[MN], use[MN], ans[MM];
pii ed[MM]; vector<int> adj[MN];
bitset<MN> ok[MN], hm;
int fnd(int x){return ds[x]=x==ds[x]?x:fnd(ds[x]);}

int main(){
    scanf("%d%d%d",&N,&M,&T);
    if(T==1){
        for(i=1;i<=M;i++) scanf("%d%d",&ed[i].first,&ed[i].second);
        for(i=1;i<=N;i++) ok[i][i]=1;
        for(i=M;i>=1;i--){
            ok[ed[i].first]|=ok[ed[i].second];
            ok[ed[i].second]|=ok[ed[i].first];
        }
        for(i=2;i<=N;i++)
            ok[1]&=ok[i];
        st = -1;
        for(i=1;i<=N;i++){
            if(ok[1][i]){
                st = i;
                break;
            }
        }
        if(st==-1) printf("-1\n");
        else{
            hm[st] = 1;
            vector<char> ans;
            for(i=M;i>=1;i--){
                if(hm[ed[i].first]) ans.pb('^');
                else ans.pb('v');
                hm[ed[i].first]=(hm[ed[i].first]||hm[ed[i].second]);
                hm[ed[i].second]=(hm[ed[i].second]||hm[ed[i].first]);
            }
            reverse(ans.begin(),ans.end());
            for(auto ch : ans) printf("%c",ch);
            printf("\n");
        }
    }
    else{
        for(i=1;i<=N;i++) ds[i]=i;
        pii ok(-1,-1);
        for(i=1;i<=M;i++){
            scanf("%d%d",&ed[i].first,&ed[i].second);
            if(!use[ed[i].first]&&!use[ed[i].second]){
                ds[fnd(ed[i].first)]=fnd(ed[i].second);
                use[ed[i].first]=use[ed[i].second]=1;
            }
            adj[ed[i].first].pb(i);
            adj[ed[i].second].pb(i);
        }
        for(i=2;i<=N;i++){
            if(fnd(i)!=fnd(1)){
                ok={1,i};
                break;
            }
        }
        if(ok.first==-1) printf("-1\n");
        else{
            tie(x,y)=ok;
            int px=0, py=0;
            while(1){
                if(px>=adj[x].size()||py>=adj[y].size()) break;
                if(adj[x][px]>adj[y][py]) swap(x,y), swap(px,py);
                if(px+1<adj[x].size()&&adj[x][px+1]==adj[y][py]){
                    int t = adj[x][px];
                    if(x==ed[t].first) ans[t]=1, x=ed[t].second;
                    else x=ed[t].first;
                    px=upper_bound(adj[x].begin(),adj[x].end(),t)-adj[x].begin();
                }
                else{
                    int t = adj[x][px];
                    if(x==ed[t].second) ans[t]=1;
                    px++;
                }
            }
            for(i=1;i<=M;i++){
                printf("%c",ans[i]?'v':'^');
            }
            printf("\n");
        }
    }
    return 0;
}