honoka和格点三角形

honoka和格点三角形
https://ac.nowcoder.com/acm/contest/3002/A

求余操作，因为1e9*1e9次方都要破ll故而每次都需要mod
两条边都平行x，y：其实就是直角三角形，考虑x为1，另一边平行于y轴；y为1时，另一边平行x轴。
sum += (n - 1) * 2 * (m - 2) * 2; 与 sum += (m - 1) * 2 * (n - 2) * 2;
只有一条边平行：如代码

#pragma warning (disable :4996)
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll modd = 1e9 + 7;
ll n, m;
ll sum = 0;
ll sum1 = 0;

int main() {
    scanf("%lld %lld", &n, &m);

        sum += ((m - 2) * (m - 2)) % modd * (((n - 2) * 2 + 2) % modd);
        sum %= modd;
        sum += ((n - 2) * (n - 2)) % modd * (((m - 2) * 2 + 2) % modd);
        sum %= modd;
        sum += ((m - 1) * (m - 2)) % modd * (n - 2) * 2;
        sum %= modd;
        sum += ((n - 1) * (n - 2)) % modd * (m - 2) * 2;
        sum %= modd;
        sum += (m - 1) * 2 * (n - 2) * 2;
        sum %= modd;
        sum += (n - 1) * 2 * (m - 2) * 2;
        sum %= modd;
        printf("%lld", sum);
        return 0;


}