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;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
      x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    return x * f;
  }
  long long cas, n, a[100005], t[2][100005], b[2][100005], tot[2], ans,
      sum[2][100005];
  long long ceil(long long x, long long y) {
    if (x % y == 0)
      return x / y;
    else
      return x / y + 1;
  }
  signed main() {
    cas = read();
    while (cas--) {
      n = read();
      for (long long i = 1; i <= n; ++i) a[i] = read();
      ans = 0;
      for (long long i = n, r = 1; i >= 1; --i, r ^= 1) {
        for (long long j = 1; j <= tot[r ^ 1]; ++j) {
          (ans +=
           t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353 +
           sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
          (sum[r ^ 1][b[r ^ 1][j]] +=
           t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353) %=
              998244353;
          long long tr = a[i] / ceil(a[i], b[r ^ 1][j]);
          if (!t[r][tr]) b[r][++tot[r]] = tr;
          (t[r][tr] += t[r ^ 1][b[r ^ 1][j]]) %= 998244353;
          (sum[r][tr] += sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
          t[r ^ 1][b[r ^ 1][j]] = 0;
          sum[r ^ 1][b[r ^ 1][j]] = 0;
        }
        tot[r ^ 1] = 0;
        if (!t[r][a[i]]) b[r][++tot[r]] = a[i];
        t[r][a[i]]++;
      }
      printf("%d\n", (ans % 998244353 + 998244353) % 998244353);
      for (long long i = 1; i <= tot[n & 1]; ++i) {
        t[n & 1][b[n & 1][i]] = 0;
        sum[n & 1][b[n & 1][i]] = 0;
      }
      tot[n & 1] = 0;
    }
    return 0;
  }
  