You are given an undirected simple connected graph with $n$ vertices and $m-1$ edges. The weights of these edges are a permutation of $2, 3, \dots, m$.

The weight of a path is defined as the product of the weights of the edges on the path. What is the maximum possible weight of a path from $s$ to $t$ such that the weight does not exceed $m$? Note that paths are allowed to revisit vertices or edges.

You need to answer multiple queries.

Input

The first line contains three positive integers $n, m, q$.

The next $m-1$ lines each contain two positive integers $u, v$, representing the two vertices connected by the edge with weight $2, 3, \dots, m$ respectively.

The next $q$ lines each contain two positive integers $s, t$, representing a query.

Output

Output $q$ lines, each containing one number. If there exists a path with a weight not exceeding $m$, output the maximum such weight; otherwise, output $-1$.

Examples

Input 1

4 5 10
1 2
2 3
3 4
4 1
1 1
1 2
1 3
1 4
2 2
2 3
2 4
3 3
3 4
4 4

Output 1

4
2
-1
5
4
3
-1
1
4
1

Subtasks

For $100\%$ of the data, it is guaranteed that $2 \leq n \leq m \leq 5 \times 10^5$ and $1 \leq q \leq 5 \times 10^5$.

For data point $1$, it is guaranteed that $m \leq 10$.

For data point $2$, it is guaranteed that $m \leq 100$.

For data points $3 \sim 4$, it is guaranteed that $m \leq 10^3$.

For data points $5 \sim 6$, it is guaranteed that $m \leq 10^4$.

For data points $7 \sim 9$, it is guaranteed that $m \leq 10^5$.

For data point $10$, there are no special restrictions.

Note

The input and output volume for this problem is large, so you may need to use I/O optimization. A template is provided below; note the location indicated in the comments for enabling file I/O:

using u32 = unsigned;
struct IO_Tp
{
    const static int _I_Buffer_Size = 30 << 20;
    char _I_Buffer[_I_Buffer_Size], *_I_pos = _I_Buffer;

    const static int _O_Buffer_Size = 8 << 20;
    char _O_Buffer[_O_Buffer_Size], *_O_pos = _O_Buffer;

    u32 m[10000];

    IO_Tp()
    {
//        freopen("tour.in", "r", stdin);
//        freopen("tour.out", "w", stdout);
        constexpr u32 e0 = '\0\0\0\1', e1 = '\0\0\1\0', e2 = '\0\1\0\0', e3 = '\1\0\0\0';
        int x = 0;
        for (u32 i = 0, c0 = '0000'; i != 10; ++i, c0 += e0)
            for (u32 j = 0, c1 = c0; j != 10; ++j, c1 += e1)
                for (u32 k = 0, c2 = c1; k != 10; ++k, c2 += e2)
                    for (u32 l = 0, c3 = c2; l != 10; ++l, c3 += e3)
                        m[x++] = c3;

        fread(_I_Buffer, 1, _I_Buffer_Size, stdin);
    }
    ~IO_Tp() { fwrite(_O_Buffer, 1, _O_pos - _O_Buffer, stdout); }

    IO_Tp &operator>>(int &res)
    {
        while (!isdigit(*_I_pos))
            ++_I_pos;
        res = *_I_pos++ - '0';
        while (isdigit(*_I_pos))
            res = res * 10 + (*_I_pos++ - '0');
        return *this;
    }

    IO_Tp &operator<<(int x)
    {
        if (x == -1)
        {
            *_O_pos++ = '-';
            *_O_pos++ = '1';
            return *this;
        }
        if (x == 0)
        {
            *_O_pos++ = '0';
            return *this; 
        }
        static char _buf[35];
        char *_pos = _buf + 35;
        while (x >= 10000)
            *--reinterpret_cast<u32 *&>(_pos) = m[x % 10000], x /= 10000;
        *--reinterpret_cast<u32 *&>(_pos) = m[x];
        _pos += (x < 1000) + (x < 100) + (x < 10);
        _O_pos = std::copy(_pos, _buf + 35, _O_pos);
        return *this;
    }

    IO_Tp &operator<<(char ch)
    {
        *_O_pos++ = ch;
        return *this;
    }
} IO;