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 ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define mp make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////

ll N,M;
ll T;
vector<pll> xyv;
ll mk[50505],mk2[50505];

ll i, j, k;
ll ans[101010];

void T1(){
    vector<bitset<50005>> bt(N+5);
    fornum(i,0,N){
        bt[i + 1][i + 1] = true;
    }
    for (i = M - 1; i >= 0;--i)
    {
        ll x = xyv[i].first;
        ll y = xyv[i].second;
        bt[y] = bt[x] = bt[x] | bt[y];
    }
    bitset<50005> btc = bt[1];
    fornum(i,1,N){
        btc &= bt[i + 1];
    }
    fornum(i,0,N){
        if(btc[i+1])
            break;
    }
    if(i==N){
        printf("-1");
        return;
    }
    ll a = i + 1;
    bitset<50005> bt2;
    bt2[a] = true;
    for (i = M - 1; i >= 0;--i)
    {
        ll x = xyv[i].first;
        ll y = xyv[i].second;
        if(bt2[y]){
            ans[i] = 1;
        }
        bt2[x] = bt2[y] = bt2[x] || bt2[y];
    }
    fornum(i,0,M){
        if(ans[i]==0){
            printf("^");
        }else{
            printf("v");
        }
    }
}


void T2(){
    if(N==2){
        printf("-1");
        return;
    }
    fornum(i,0,N){
        mk[i + 1] = i + 1;
        mk2[i + 1] = 1;
    }
    for (i = M - 1; i >= 0;--i){
        ll x = xyv[i].first;
        ll y = xyv[i].second;
        if(mk2[mk[x]]>1){
            ans[i] = 1;
            --mk2[mk[x]];
            ++mk2[mk[y]];
            mk[x] = mk[y];
        }else{
            ++mk2[mk[x]];
            --mk2[mk[y]];
            mk[y] = mk[x];
        }
    }
    fornum(i,0,M){
        if(ans[i]==0){
            printf("^");
        }else{
            printf("v");
        }
    }
}




int main(){
    scanf("%lld%lld%lld", &N,&M,&T);
    fornum(i,0,M){
        ll x, y;
        scanf("%lld%lld", &x, &y);
        xyv.push_back({x, y});
    }

    if(T==1){
        T1();
    }else{
        T2();
    }
    
    
    
    return 0;
}