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>
  using namespace std;
  namespace Legendgod {
  namespace Read {
  template <typename T>
  void r1(T &x) {
    x = 0;
    int f(1);
    char c(getchar());
    for (; !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) x = (x * 10) + (c ^ 48);
    x *= f;
  }
  template <typename T, typename... Args>
  void r1(T &x, Args &...arg) {
    r1(x), r1(arg...);
  }
  }  // namespace Read
  using namespace Read;
  const int maxn = 2e5 + 5;
  const int mod = 998244353;
  int n, a[maxn];
  int f[2][maxn];
  vector<int> buc[2];
  void Solve() {
    int i, j;
    r1(n);
    for (i = 1; i <= n; ++i) r1(a[i]);
    for (int v : buc[0]) f[0][v] = 0;
    for (int v : buc[1]) f[1][v] = 0;
    buc[0].clear(), buc[1].clear();
    int ans(0);
    for (i = n; i >= 1; --i) {
      int now = (i & 1), bef(now ^ 1);
      for (int v : buc[now]) f[now][v] = 0;
      buc[now].clear();
      buc[now].push_back(a[i]);
      f[now][a[i]] = 1;
      int last = a[i];
      for (int v : buc[bef]) {
        int K = (a[i] + v - 1) / v;
        int st = a[i] / K;
        ans = (ans + 1ll * (K - 1) * i % mod * f[bef][v] % mod) % mod;
        f[now][st] = (f[now][st] + f[bef][v]) % mod;
        if (last != st) buc[now].push_back(st), last = st;
      }
    }
    printf("%d\n", ans);
  }
  signed main() {
    int i, j, T;
    r1(T);
    while (T--) Solve();
    return 0;
  }
  }  // namespace Legendgod
  signed main() { return Legendgod::main(), 0; }
  