func_code_string: #include <bits/stdc++.h>   #pragma GCC optimize(2)   #pragma GCC optimize(3)   using namespace std;   template <typename T1, typename T2>   void ckmin(T1 &a, T2 b) {     if (a > b) a = b;   }   template <typename T1, typename T2>   void ckmax(T1 &a, T2 b) {     if (a < b) a = b;   }   template <typename T>   T gcd(T a, T b) {     return b ? gcd(b, a % b) : a;   }   int read() {     int x = 0, f = 0;     char ch = getchar();     while (!isdigit(ch)) f |= ch ==  - , ch = getchar();     while (isdigit(ch)) x = 10 * x + ch -  0 , ch = getchar();     return f ? -x : x;   }   const int mod = 1000000007;   const int N = 2005;   struct fuck {     int x, y, b;   } a[N];   int sum[N][N];   int n, m, R, t = 1000;   int fac[N], ifac[N];   int qpow(int a, int b = mod - 2) {     int res = 1;     while (b > 0) {       if (b & 1) res = 1ll * res * a % mod;       a = 1ll * a * a % mod;       b >>= 1;     }     return res;   }   void init(int n) {     fac[0] = ifac[0] = 1;     for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;     ifac[n] = qpow(fac[n], mod - 2);     for (int i = n - 1; i >= 1; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;   }   int C(int n, int m) {     if (n < m || n < 0)       return 0;     else       return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod;   }   int calc1(int xd1, int yd1, int xd2, int yd2) {     return sum[xd2][yd2] - sum[xd1 - 1][yd2] - sum[xd2][yd1 - 1] +            sum[xd1 - 1][yd1 - 1];   }   int main() {     init(2000);     n = read(), m = read(), R = read();     for (int i = 1; i <= n; i++) {       a[i].x = read(), a[i].y = read(), a[i].b = read();       sum[a[i].x][a[i].y]++;     }     for (int i = 1; i <= t; i++) {       for (int j = 1; j <= t; j++) {         sum[i][j] += sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1];       }     }     long long ans = 0;     for (int i = 1; i <= n; i++) {       int xd1 = max(1, a[i].x - R), yd1 = max(1, a[i].y - R);       int xd2 = min(t, a[i].x + R), yd2 = min(t, a[i].y + R);       int in_dot = calc1(xd1, yd1, xd2, yd2);       ans = (ans + 1ll * (C(n, m) - C(n - in_dot, m) + mod) * a[i].b % mod *                        a[i].b % mod) %             mod;     }     for (int i = 1; i <= n; i++) {       for (int j = i + 1; j <= n; j++) {         int xd1 = max(1, a[i].x - R), yd1 = max(1, a[i].y - R);         int xd2 = min(t, a[i].x + R), yd2 = min(t, a[i].y + R);         int in_dot = calc1(xd1, yd1, xd2, yd2);         int kase = C(n, m);         kase = (kase - C(n - in_dot, m) + mod) % mod;         int x3 = max(1, a[j].x - R), y3 = max(1, a[j].y - R);         int x4 = min(t, a[j].x + R), y4 = min(t, a[j].y + R);         int in_dot2 = calc1(x3, y3, x4, y4);         kase = (kase - C(n - in_dot2, m) + mod) % mod;         int x5 = max(xd1, x3), y5 = max(yd1, y3);         int x6 = min(xd2, x4), y6 = min(yd2, y4);         int in_dot3 = 0;         if (x5 <= x6 && y5 <= y6) {           in_dot3 = calc1(x5, y5, x6, y6);         }         int in = in_dot + in_dot2 - in_dot3;         kase = (kase + C(n - in, m)) % mod;         ans = (ans + 1ll * 2 * a[i].b % mod * a[j].b % mod * kase % mod) % mod;       }     }     printf( %lld n , (ans % mod + mod) % mod);     return 0;   }   