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 long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vp;
const int inf = 1e9;
const ll INF = 1e18;
const db eps = 1e-10;

#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define ep emplace
#define mem(a) memset(a, 0, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(b))
#define PA cout << "pass\n"
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define TM cout << db(clock()) / CLOCKS_PER_SEC << '\n'

const int maxn = 5e4 + 233;
bitset<maxn> mp[maxn];
int n, m, T;
int x[maxn * 2], y[maxn * 2];
bool vis[maxn];
char s[maxn * 2];
int cnt[maxn], pos[maxn];

int main(){
    scanf("%d %d %d", &n, &m, &T);
    if(T == 1) {
        for (int i = 1; i <= n; i++) mp[i].set(i);
        for (int i = 1; i <= m; i++) {
            scanf("%d %d", &x[i], &y[i]);
            mp[x[i]] = mp[y[i]] = mp[x[i]] | mp[y[i]];
        }
        for (int i = 1; i <= n; i++) {
            if (mp[i].count() == n) {
                vis[i] = 1;
                for (int j = m; j; j--) {
                    if (vis[x[j]]) {
                        s[j] = '^';
                        vis[y[j]] = 1;
                    } else if (vis[y[j]]) {
                        vis[x[j]] = 1;
                        s[j] = 'v';
                    } else s[j] = 'v';
                }
                for (int j = 1; j <= m; j++) printf("%c", s[j]);
                puts("");
                return 0;
            }
        }
        puts("-1");
    }
    else{
        for(int i = 1; i <= m; i++) scanf("%d %d", &x[i], &y[i]);
        if(n == 2) puts("-1");
        else {
            for (int i = 1; i <= n; i++) cnt[i] = 1, pos[i] = i;
            for (int i = m; i; i--) {
                if (cnt[pos[x[i]]] >= 2) {
                    cnt[pos[x[i]]]--;
                    pos[x[i]] = pos[y[i]];
                    cnt[pos[y[i]]]++;
                    s[i] = 'v';
                } else {
                    cnt[pos[y[i]]]--;
                    pos[y[i]] = pos[x[i]];
                    cnt[pos[x[i]]]++;
                    s[i] = '^';
                }
            }
            s[m + 1] = '\0';
            puts(s + 1);
        }
    }
    return 0;
}