problem

E2. Close Tuples (hard version)
time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
This is the hard version of this problem. The only difference between the easy and hard versions is the constraints on k and m. In this version of the problem, you need to output the answer by modulo 109+7.

You are given a sequence a of length n consisting of integers from 1 to n. The sequence may contain duplicates (i.e. some elements can be equal).

Find the number of tuples of m elements such that the maximum number in the tuple differs from the minimum by no more than k. Formally, you need to find the number of tuples of m indices i1<i2<…<im, such that

max(ai1,ai2,…,aim)−min(ai1,ai2,…,aim)≤k.
For example, if n=4, m=3, k=2, a=[1,2,4,3], then there are two such triples (i=1,j=2,z=4 and i=2,j=3,z=4). If n=4, m=2, k=1, a=[1,1,1,1], then all six possible pairs are suitable.

As the result can be very large, you should print the value modulo 109+7 (the remainder when divided by 109+7).

Input
The first line contains a single integer t (1≤t≤2⋅105) — the number of test cases. Then t test cases follow.

The first line of each test case contains three integers n, m, k (1≤n≤2⋅105, 1≤m≤100, 1≤k≤n) — the length of the sequence a, number of elements in the tuples and the maximum difference of elements in the tuple.

The next line contains n integers a1,a2,…,an (1≤ai≤n) — the sequence a.

It is guaranteed that the sum of n for all test cases does not exceed 2⋅105.

Output
Output t answers to the given test cases. Each answer is the required number of tuples of m elements modulo 109+7, such that the maximum value in the tuple differs from the minimum by no more than k.

Example
inputCopy
4
4 3 2
1 2 4 3
4 2 1
1 1 1 1
1 1 1
1
10 4 3
5 6 1 3 2 9 8 1 2 4
outputCopy
2
6
1
20

solution

/*
题意：
+ 给出一个长为n的序列，求有多少个m=3元组（任选m个元素组成的集合）满足其中最大数-最小数小于等于k=2。
思路：
+ 先对数组排序，然后算出每个数的贡献，先从第一个数开始找到第一个大于它的值+k的数（二分），下标差即为可选的构成m个数元组的可选数的个数s。
+ 如果s<m贡献0，如果s>=m,贡献为从s-1个数中选出m-1个数的方法数（已经选了一个数），转换为求组合数的模问题。
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 5;
const int mod = 1e9+7;
LL inv(LL x){return x==1?1:(LL)(mod-mod/x)*inv(mod%x)%mod;}
LL C(LL n, LL m){
  if(n<0 || n<m)return 0;
  if(m>n-m)m=n-m;
  LL up = 1, down = 1;
  for(LL i = 0; i < m; i++){
    up = up*(n-i)%mod;
    down = down*(i+1)%mod;
  }
  return up*inv(down)%mod;
}
int a[maxn];
int main(){
  int T;  cin>>T;
  while(T--){
    int n, m, k;  cin>>n>>m>>k;
    //int n, m=3, k=2;  cin>>n;
    for(int i = 1; i <= n; i++)cin>>a[i];
    sort(a+1,a+n+1);
    LL ans = 0;
    for(int i = 1; i <= n; i++){
      int r = upper_bound(a+1,a+n+1,a[i]+k)-a;
      int cnt = r-i;
      if(cnt<m)continue;
      ans += C(cnt-1,m-1);
      ans %= mod;
    }
    cout<<ans<<"\n";
  }
  return 0;
}