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 <cstdio>
#include <algorithm>
#include <vector>
#include <set>
#include <cstdlib>
#include <utility>
#include <cmath>
#include <queue>
#include <stack>
#include <cstring>

using namespace std;

#define ll long long

#ifndef ONLINE_JUDGE
#define debug(format, ...) fprintf(stderr, \
    "%s:%d: " format "\n", __func__, __LINE__,##__VA_ARGS__)
#else
#define debug(format, ...)
#define NDEBUG
#endif

int n, m, T;
int x[100005], y[100005];
char ans[100005];

struct Edge {
    int from; int to; int id;
    unsigned int* mask;
};

vector<Edge> graph[50005];

void Or(unsigned int* res, unsigned int* a, unsigned int* b)
{
    for (int i = 0; i < 1600; i++) res[i] = a[i] | b[i];
}

Edge edges[100005];

unsigned int zero[1600];

unsigned int* mask_above(int node, int id_above)
{
    if (graph[node].back().id <= id_above) return zero;
    int l = 0; int r = (int)(graph[node].size()) - 1;
    while (l < r) {
        int mid = (l + r) / 2;
        if (graph[node][mid].id > id_above) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return graph[node][l].mask;
}

unsigned int curmask[1600];
int f[50005];

void solve_1()
{
    for (int i = m; i >= 1; i--) {
        Edge e; e.from = x[i]; e.to = y[i]; e.id = m - i + 1;
        e.mask = new unsigned int[1600];
        graph[x[i]].push_back(e);
        e.from = y[i]; e.to = x[i];
        graph[y[i]].push_back(e);
        edges[m - i + 1] = e;
    }
    for (int i = m; i >= 1; i--) {
        int from = edges[i].from;
        int to = edges[i].to;
        unsigned int* mask_from = mask_above(from, i);
        unsigned int* mask_to = mask_above(to, i);
        Or(edges[i].mask, mask_from, mask_to);
        edges[i].mask[(from - 1) / 32] |= (1U << (unsigned int)((from - 1) % 32));
        edges[i].mask[(to - 1) / 32] |= (1U << (unsigned int)((to - 1) % 32));

    }
    int start = -1;
    for (int i = 1; i <= n; i++) {
        if (!(graph[i].size())) continue;
        for (int j = 0; j < 1600; j++) curmask[j] = graph[i][0].mask[j];
        bool good = true;
        for (int j = 0; j < n / 32; j++) {
            if (curmask[j] != (~(0U))) good = false;
        }
        for (int j = n / 32 * 32; j < n; j++) {
            if (!(curmask[n / 32] & (1U << (unsigned int)(j % 32)))) good = false;
        }
        if (good) start = i;
    }
    if (start == -1) {
        printf("-1"); exit(0);
    } else {
        //printf("%d\n", start);
        /*#ifndef ONLINE_JUDGE
        start = 3;
        #endif*/
        for (int i = 1; i <= n; i++) f[i] = i;
        for (int i = m; i >= 1; i--) {
            if (f[x[i]] == start) {
                ans[i] = '^'; f[y[i]] = f[x[i]];
            } else {
                ans[i] = 'v'; f[x[i]] = f[y[i]];
            }
        }
    }
}

int freq[50005];

void solve_2()
{
    if (n == 2) {
        printf("-1"); exit(0);
    }
    for (int i = 1; i <= n; i++) f[i] = i;
    for (int i = 1; i <= n; i++) freq[i] = 1;
    for (int i = m; i >= 1; i--) {
        if (freq[f[x[i]]] == n - 1) {
            ans[i] = 'v';
        } else {
            ans[i] = '^';
        }
        if (ans[i] == '^') {
            freq[f[y[i]]]--; freq[f[x[i]]]++; f[y[i]] = f[x[i]];
        } else {
            freq[f[x[i]]]--; freq[f[y[i]]]++; f[x[i]] = f[y[i]];
        }
    }
}

int main()
{
    scanf("%d%d%d", &n, &m, &T);
    for (int i = 1; i <= m; i++) scanf("%d%d", x + i, y + i);
    if (T == 1) {
        solve_1();
    } else {
        solve_2();
    }
    for (int i = 1; i <= m; i++) printf("%c", ans[i]);
    return 0;
}