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>
  const int N = 1e5 + 5;
  const int K = 655;
  const int mod = 998244353;
  int T;
  int n, a[N];
  int q[K], v[K], w[K], top = 0;
  int q2[K], v2[K], w2[K], top2 = 0;
  int MOD(int x) { return x >= mod ? x - mod : x; }
  int u(int x, int y) { return (x + y - 1) / y; }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
      q[top = 1] = a[n], v[1] = 0, w[1] = 1;
      int ans = 0;
      for (int i = n - 1; i >= 1; --i) {
        top2 = 0;
        for (int l = 1, r, j = 0; l <= a[i]; l = r + 1) {
          r = a[i] / (a[i] / l);
          q2[++top2] = a[i] / l;
          v2[top2] = 0;
          w2[top2] = (l == 1);
          if (i == 1) {
            int tmp = 1;
          }
          while (j < top && u(a[i], q[j + 1]) >= l && u(a[i], q[j + 1]) <= r) {
            ++j;
            v2[top2] = MOD(v2[top2] + MOD(v[j] + (long long)(u(a[i], q[j]) - 1) *
                                                     w[j] % mod));
            w2[top2] = w2[top2] + w[j];
          }
        }
        for (int j = 1; j <= top2; ++j) {
          q[j] = q2[j];
          v[j] = v2[j];
          w[j] = w2[j];
        }
        top = top2;
        for (int j = 1; j <= top; ++j) ans = MOD(ans + v[j]);
      }
      printf("%d\n", ans);
    }
    return 0;
  }
  