description: For an array b of n integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make b non-decreasing: 
  
    * Select an index i such that 1 ≤ i ≤ |b|, where |b| is the current length of b. 
    * Replace b_i with two elements x and y such that x and y both are positive integers and x + y = b_i. 
    * This way, the array b changes and the next operation is performed on this modified array. 
  
  
  
  For example, if b = [2, 4, 3] and index 2 gets selected, then the possible arrays after this operation are [2, \underline{1}, \underline{3}, 3], [2, \underline{2}, \underline{2}, 3], or [2, \underline{3}, \underline{1}, 3]. And consequently, for this array, this single operation is enough to make it non-decreasing: [2, 4, 3] → [2, \underline{2}, \underline{2}, 3].
  
  It's easy to see that every array of positive integers can be made non-decreasing this way.
  
  YouKn0wWho has an array a of n integers. Help him find the sum of extreme values of all nonempty subarrays of a modulo 998 244 353. If a subarray appears in a multiple times, its extreme value should be counted the number of times it appears.
  
  An array d is a subarray of an array c if d can be obtained from c by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^5).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
  
  Output
  
  For each test case, print a single integer — the sum of extreme values of all subarrays of a modulo 998 244 353.
  
  Example
  
  Input
  
  
  4
  3
  5 4 3
  4
  3 2 1 4
  1
  69
  8
  7264 40515 28226 92776 35285 21709 75124 48163
  
  
  Output
  
  
  5
  9
  0
  117
  
  Note
  
  Let f(l, r) denote the extreme value of [a_l, a_{l+1}, …, a_r].
  
  In the first test case, 
  
    * f(1, 3) = 3, because YouKn0wWho can perform the following operations on the subarray [5, 4, 3] (the newly inserted elements are underlined):
  
  [5, 4, 3] → [\underline{3}, \underline{2}, 4, 3] → [3, 2, \underline{2}, \underline{2}, 3] → [\underline{1}, \underline{2}, 2, 2, 2, 3];
  
    * f(1, 2) = 1, because [5, 4] → [\underline{2}, \underline{3}, 4];
    * f(2, 3) = 1, because [4, 3] → [\underline{1}, \underline{3}, 3];
    * f(1, 1) = f(2, 2) = f(3, 3) = 0, because they are already non-decreasing. 
  
  
  
  So the total sum of extreme values of all subarrays of a = 3 + 1 + 1 + 0 + 0 + 0 = 5.
solution: #include <bits/stdc++.h>
  #pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                       "omit-frame-pointer", "inline")
  #pragma GCC option("arch=native", "tune=native", "no-zero-upper")
  using namespace std;
  const int MAXN = 1e5 + 100;
  int A[MAXN];
  const int MOD = 998244353;
  int mp1[MAXN];
  int mp2[MAXN];
  bool mk1[MAXN];
  bool mk2[MAXN];
  vector<int> v1, v2;
  void limpiar1() {
    for (auto x : v1) {
      mk1[x] = false;
      mp1[x] = 0;
    }
    v1.clear();
  }
  void limpiar2() {
    for (auto x : v2) {
      mk2[x] = false;
      mp2[x] = 0;
    }
    v2.clear();
  }
  void sumar1(int x, int v) {
    if (!mk1[x]) mk1[x] = true, v1.push_back(x);
    mp1[x] += v;
  }
  void sumar2(int x, int v) {
    if (!mk2[x]) mk2[x] = true, v2.push_back(x);
    mp2[x] += v;
  }
  void copiar() {
    for (auto x : v2) {
      sumar1(x, mp2[x]);
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int N;
      cin >> N;
      for (int i = 1; i <= N; i++) {
        cin >> A[i];
      }
      int res = 0;
      for (int i = N; i >= 1; i--) {
        for (auto x : v1) {
          if (A[i] <= x) {
            sumar2(A[i], mp1[x]);
          } else if (A[i] % x == 0) {
            sumar2(x, mp1[x]);
            res += (long long)(A[i] / x - 1) * (long long)mp1[x] % MOD *
                   (long long)i % MOD;
            res %= MOD;
          } else {
            sumar2(A[i] / (A[i] / x + 1), mp1[x]);
            res += (long long)(A[i] / x) * (long long)mp1[x] % MOD *
                   (long long)i % MOD;
            res %= MOD;
          }
        }
        sumar2(A[i], 1);
        limpiar1();
        copiar();
        limpiar2();
      }
      limpiar1();
      cout << res << '\n';
    }
    return 0;
  }
  