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>
  #pragma GCC optimize("Ofast,unroll-loops,inline")
  #pragma GCC option("arch=native", "tune=native", "no-zero-upper")
  #pragma GCC target( \
      "bmi,bmi2,lzcnt,popcnt,movbe,aes,pclmul,rdrnd,abm,mmx,avx,avx2,f16c,fma,sse,sse2,sse3,ssse3,sse4.1,sse4.2")
  using namespace std;
  using ll = long long;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vii = vector<pii>;
  using vll = vector<pll>;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long double eps = 1e-9;
  const long long mod = 998244353;
  const int MAXN = 200000;
  void solve() {
    ll n, t = 0;
    cin >> n;
    int arr[n];
    for (int i = 0, nnnn = (n); i < nnnn; ++i) cin >> arr[i];
    vector<pair<int, ll>> cur, nxt, nxt1;
    for (int i = n - 1; i >= 0; i--) {
      nxt.clear();
      nxt1.clear();
      for (auto it : cur) {
        t += (it.first - 1) * it.second * (i + 1);
        if (i > 0) {
          int t1 = arr[i] / it.first;
          int add = (arr[i - 1] + t1 - 1) / t1;
          nxt.push_back(make_pair(add, it.second));
        }
      }
      t %= mod;
      if (i > 0) {
        ll d = arr[i - 1] / arr[i] + (int)(arr[i - 1] % arr[i] > 0);
        nxt.push_back(make_pair(d, 1));
      }
      sort((nxt).begin(), (nxt).end());
      for (auto it : nxt) {
        if (nxt1.empty())
          nxt1.push_back(it);
        else if (it.first == nxt1[((int)(nxt1).size()) - 1].first) {
          nxt1[((int)(nxt1).size()) - 1].second += it.second;
        } else
          nxt1.push_back(it);
      }
      cur = nxt1;
    }
    cout << t << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(12) << fixed;
    int t;
    cin >> t;
    while (t--) solve();
  }
  