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;
  const int p = 998244353;
  const int N = 1e5 + 5;
  inline int inc(int x, int y) {
    if ((x += y) >= p) x -= p;
    return x;
  }
  inline void Inc(int &x, int y) { x = inc(x, y); }
  inline int dec(int x, int y) {
    if ((x -= y) < 0) x += p;
    return x;
  }
  inline void Dec(int &x, int y) { x = dec(x, y); }
  inline int ksm(int a, int b) {
    int ans = 1;
    for (; b; b >>= 1, a = 1ll * a * a % p)
      if (b & 1) ans = 1ll * a * ans % p;
    return ans;
  }
  int n, a[N], b[N], mx[N], Ans, ans, sum;
  inline void solve() {
    scanf("%d", &n);
    ans = 0;
    Ans = 0;
    sum = 0;
    for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
    mx[n + 1] = p;
    for (int i = (n); i >= (1); i--) {
      if (a[i] <= mx[i + 1])
        mx[i] = a[i], b[i] = 1;
      else {
        b[i] = (a[i] / mx[i + 1]) + (a[i] % mx[i + 1] != 0);
        mx[i] = (a[i] / b[i]);
      }
      Inc(sum, b[i] - 1);
      Inc(ans, sum);
    }
    Ans = ans;
    for (int i = (n - 1); i >= (1); i--) {
      Dec(ans, 1ll * i * (b[i] - 1) % p);
      if (a[i] != mx[i]) {
        mx[i] = a[i];
        b[i] = 1;
        for (int j = (i - 1); j >= (1); j--) {
          int lstmx = mx[j];
          Dec(ans, 1ll * j * (b[j] - 1) % p);
          if (a[j] <= mx[j + 1])
            mx[j] = a[j], b[j] = 1;
          else {
            b[j] = (a[j] / mx[j + 1]) + (a[j] % mx[j + 1] != 0);
            mx[j] = (a[j] / b[j]);
          }
          Inc(ans, 1ll * j * (b[j] - 1) % p);
          if (mx[j] == lstmx) break;
        }
      }
      Inc(Ans, ans);
    }
    printf("%d\n", Ans);
  }
  int main() {
    int T;
    cin >> T;
    while (T--) solve();
  }
  