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() {
    register long long a = 0, po = 1;
    char ch = getchar();
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') po = -1, ch = getchar();
    while (isdigit(ch)) a = (a << 1) + (a << 3) + ch - 48, ch = getchar();
    return a * po;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
  }
  const long long maxn = 100010;
  const long long mod = 998244353;
  struct nowt {
    long long t, cy, cc;
  };
  void work() {
    vector<nowt> N, N1;
    long long n = read();
    long long tot = 0;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      a[i] = read();
    }
    N1.push_back({1, a[n - 1], n});
    for (long long i = n - 2; i >= 0; i--) {
      N.push_back({1, a[i], i + 1});
      long long lenn = (long long)N1.size();
      for (long long j = 0; j < (long long)N1.size(); j++) {
        long long t = (a[i] + N1[j].cy - 1) / N1[j].cy;
        long long d = a[i] / t;
        long long num = N.size();
        if (t != N[num - 1].t || d != N[num - 1].cy)
          N.push_back({t, d, N1[j].cc});
      }
      N.push_back({0, 0, n + 1});
      long long curr = (long long)N.size();
      for (long long j = 0; j < (long long)N.size() - 1; j++) {
        tot %= mod;
        tot += (i + 1) * (N[j].t - 1) % mod * (N[j + 1].cc - N[j].cc) % mod;
        tot %= mod;
      }
      N.pop_back();
      N1 = N;
      N.clear();
    }
    cout << tot << endl;
    a.clear();
  }
  signed main() {
    long long tt = read();
    while (tt--) {
      work();
    }
  }
  