CF894B：Ralph And His Magic Field（思维）_c++[cf894b] ralph and his magic field-优快云博客

探讨如何用1和-1填满n*m的矩阵，确保每一行每一列的乘积为k，输出方案数并模1e9+7。通过分析矩阵填充规律，利用快速幂求解。

Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.

Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.

Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.

Input

The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).

Output

Print a single number denoting the answer modulo 1000000007.

Examples

input
1 1 -1

output
1

input
1 3 1

output
1

input
3 3 -1

output
16

Note

In the first example the only way is to put -1 into the only block.

In the second example the only way is to put 1 into every block.

题意：用1和-1填满n*m的矩阵，令到每一行每一列的乘积为k，输出方案数mod 1e9+7。

思路：先不考虑最后一行和最后一列，任意填充前(n-1)*(m-1)个格，那么最后一行和最后一列显然可以填上1或-1使得每行每列都等于k，那么右下角那个格子呢？设前(n-1)*(m-1)个格乘积为sum，对于列来说，右下角填上A=k^(n-1)*sum，对于行来说右下角填上B=k^(m-1)*sum。那么显然如果k=-1，且n和m奇偶性不同时A不等于B，其余情况都能满足。答案就是pow(2, (n-1)*(m-1))，范围太大用欧拉降幂。

# include <iostream>
# include <cstdio>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
LL qmod(LL a, LL b)
{
    LL ans = 1;
    for(;b;b>>=1)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
    }
    return ans;
}
int main()
{
    LL n, m, k;
    scanf("%lld%lld%lld",&n,&m,&k);
    if(k == -1 && (n&1)!=(m&1)) return 0*puts("0");
    n = (n-1)%(mod-1);
    m = (m-1)%(mod-1);
    printf("%lld\n",qmod(2LL,n*m%(mod-1)+mod-1));
    return 0;
}