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:
/*
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*/

#define _CRT_SECURE_NO_WARNINGS

#include <map>
#include <set>
#include <stack>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cctype>
#include <vector>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <fstream>
#include <iostream>
#include <algorithm>

using namespace std;

inline char gc() {
    static const int L = 233333;
    static char sxd[L], *sss = sxd, *ttt = sxd;
    if (sss == ttt) {
        ttt = (sss = sxd) + fread(sxd, 1, L, stdin);
        if (sss == ttt) {
            return EOF;
        }
    }
    return *sss++;
}

#ifndef _AT_HOME
#define dd c = gc()
#else
#define dd c = getchar()
#endif
inline char readalpha() {
    char dd;
    for (; !isalpha(c); dd);
    return c;
}

inline char readchar() {
    char dd;
    for (; c == ' '; dd);
    return c;
}

template <class T>
inline bool read(T& x) {
    bool flg = false;
    char dd;
    x = 0;
    for (; !isdigit(c); dd) {
        if (c == '-') {
            flg = true;
        } else if(c == EOF) {
            return false;
        }
    }
    for (; isdigit(c); dd) {
        x = (x * 10) + (c ^ 48);
    }
    if (flg) {
        x = -x;
    }
    return true;
}
#undef dd

template <class T>
inline void write(T x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x < 10) {
        putchar(x | 48);
        return;
    }
    write(x / 10);
    putchar((x % 10) | 48);
}

template <class T>
inline void writesp(T x) {
    write(x);
    putchar(' ');
}

template <class T>
inline void writeln(T x) {
    write(x);
    puts("");
}

#define lowbit(x) (x & -x)

namespace dfcmd {
    const int maxn = 50005;
    const int maxm = 100005;

    int n, m, T;

    struct Balancer {
        int x, y;
    } bal[maxm];

    char ans[maxm];
    int ff[maxn];

    namespace solve1 {
        bitset<maxn> f[maxn];

        inline void solve() {
            for (int i = 1; i <= n; ++i) {
                f[i][i] = 1;
            }
            for (int i = m; i; --i) {
                f[bal[i].x] = f[bal[i].y] = f[bal[i].x] | f[bal[i].y];
            }
            for (int i = 2; i <= n; ++i) {
                f[1] &= f[i];
            }
            int now = 0;
            for (int i = 1; i <= n; ++i) {
                if (f[1][i]) {
                    now = i;
                    break;
                }
            }
            if (!now) {
                puts("-1");
                return;
            }
            ff[now] = 1;
            for (int i = m; i; --i) {
                if (ff[bal[i].x]) {
                    ans[i] = '^';
                } else {
                    ans[i] = 'v';
                }
                ff[bal[i].x] = ff[bal[i].y] = ff[bal[i].x] | ff[bal[i].y];
            }
            puts(ans + 1);
        }
    }

    namespace solve2 {
        int g[maxn];

        inline void solve() {
            if (n == 2) {
                puts("-1");
                return;
            }
            for (int i = 1; i <= n; ++i) {
                ff[i] = i;
                g[i] = 1;
            }
            for (int i = m; i; --i) {
                if (g[ff[bal[i].x]] < g[ff[bal[i].y]]) {
                    g[ff[bal[i].x]]++, g[ff[bal[i].y]]--;
                    ff[bal[i].y] = ff[bal[i].x];
                    ans[i] = '^';
                } else {
                    g[ff[bal[i].x]]--, g[ff[bal[i].y]]++;
                    ff[bal[i].x] = ff[bal[i].y];
                    ans[i] = 'v';
                }
            }
            puts(ans + 1);
        }
    }

    int Main() {
        read(n), read(m), read(T);
        for (int i = 1; i <= m; ++i) {
            read(bal[i].x), read(bal[i].y);
        }
        if (T == 1) {
            solve1::solve();
        } else {
            solve2::solve();
        }
        return 0;
    }
}

int main() {
    return dfcmd::Main();
}