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;
  template <typename T>
  void Print(T v) {
    for (auto i : v) cout << i << (" ");
    cout << '\n';
  }
  int INF = 1000000007;
  long long LINF = 1000000000000000000;
  long long mod1 = 998244353;
  long long dp[2][100001];
  long long ceil(int a, int b) {
    long long res = a / b;
    if (a % b) res++;
    return res;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> vec(n);
    for (int i = 0; i < n; i++) cin >> vec[i];
    long long res = 0;
    dp[1][vec[n - 1]] = 1;
    vector<vector<long long>> iter(2);
    iter[1].push_back(vec[n - 1]);
    for (int i = n - 2; i >= 0; i--) {
      iter[0].push_back(vec[i]);
      dp[0][vec[i]] = 1;
      while (not iter[1].empty()) {
        long long pod = ceil(vec[i], iter[1].back());
        if (dp[0][vec[i] / pod] == 0) iter[0].push_back(vec[i] / pod);
        dp[0][vec[i] / pod] += dp[1][iter[1].back()];
        res += (i + 1) * dp[1][iter[1].back()] * (pod - 1);
        res %= mod1;
        dp[1][iter[1].back()] = 0;
        iter[1].pop_back();
      }
      for (auto j : iter[0]) swap(dp[1][j], dp[0][j]);
      swap(iter[0], iter[1]);
    }
    while (not iter[1].empty()) {
      dp[1][iter[1].back()] = 0;
      iter[1].pop_back();
    }
    cout << (res) << '\n';
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
  }
  