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;

constexpr int kN = 50000;

int main() {
    int n, m, t; scanf("%d%d%d", &n, &m, &t);
    if (t == 1) {
        vector<bitset<kN>> bs(n);
        for (int i = 0; i < n; ++i) bs[i].set(i);
        vector<pair<int, int>> ed(m);
        for (int i = 0; i < m; ++i) {
            int u, v; scanf("%d%d", &u, &v);
            u--, v--;
            bs[u] |= bs[v];
            bs[v] |= bs[u];
            ed[i] = make_pair(u, v);
        }
        for (int i = 0; i < n; ++i) {
            if (bs[i].count() == n) {
                vector<int> ok(n);
                ok[i] = 1;
                bool fd = false;
                string ans(m, '^');
                for (int j = m - 1; j >= 0; --j) {
                    fd |= (ed[j].first == i || ed[j].second == i);
                    if (!fd) continue;
                    if (!ok[ed[j].first] && !ok[ed[j].second]) continue;
                    if (ok[ed[j].first] && ok[ed[j].second]) continue;
                    if (ok[ed[j].second]) ans[j] = 'v';
                    ok[ed[j].first] = ok[ed[j].second] = 1;
                }
                for (int j = 0; j < n; ++j) assert(ok[j]);
                printf("%s\n", ans.c_str());
                return 0;
            }
        }
        puts("-1");
    } else {
        if (n == 2) {
            puts("-1");
            return 0;
        }
        vector<pair<int, int>> ed(m);
        for (int i = 0; i < m; ++i) {
            scanf("%d%d", &ed[i].first, &ed[i].second);
            ed[i].first--;
            ed[i].second--;
        }
        vector<int> dest(n), cnt(n);
        for (int i = 0; i < n; ++i) dest[i] = i, cnt[i] = 1;
        string ans(m, '^');
        for (int i = m - 1; i >= 0; --i) {
            if (dest[ed[i].first] == dest[ed[i].second]) continue;
            if (cnt[dest[ed[i].second]] == n - 1) {
                --cnt[dest[ed[i].second]];
                dest[ed[i].second] = dest[ed[i].first];
                ++cnt[dest[ed[i].second]];
            } else {
                --cnt[dest[ed[i].first]];
                dest[ed[i].first] = dest[ed[i].second];
                ++cnt[dest[ed[i].first]];
                ans[i] = 'v';
            }
        }
        printf("%s\n", ans.c_str());
    }
    return 0;
}