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;

const int maxn = 100010;
int n, m, tp, x[maxn], y[maxn], c[maxn], tr[maxn];
bitset<50010> b[50010];
bool chk[maxn];
char s[maxn];

int main() {
    scanf("%d %d %d", &n, &m, &tp);
    for (int i = 1; i <= m; i++) {
        scanf("%d %d", &x[i], &y[i]);
    }
    auto subtask1 = [&]() {
        for (int i = 1; i <= n; i++) {
            b[i].set(i);
        }
        for (int i = 1; i <= m; i++) {
            b[x[i]] = b[y[i]] = b[x[i]] | b[y[i]];
        }
        for (int i = 1; i <= n; i++) if (b[i].count() == n) {
            chk[i] = 1;
            for (int j = m; j; j--) {
                if (chk[x[j]]) {
                    s[j] = '^', chk[y[j]] = 1;
                } else {
                    s[j] = 'v', chk[x[j]] |= chk[y[j]];
                }
            }
            printf("%s\n", s + 1); return;
        }
        printf("-1\n");
    };
    auto subtask2 = [&]() {
        if (n == 2) { printf("-1\n"); return; }
        for (int i = 1; i <= n; i++) {
            tr[i] = i, c[i] = 1;
        }
        for (int i = m; i; i--) {
            if (c[tr[x[i]]] >= 2) {
                s[i] = 'v';
                c[tr[x[i]]]--, c[tr[x[i]] = tr[y[i]]]++;
            } else {
                s[i] = '^';
                c[tr[y[i]]]--, c[tr[y[i]] = tr[x[i]]]++;
            }
        }
        printf("%s\n", s + 1);
    };
    if (tp == 1) subtask1();
    else subtask2();
    return 0;
}