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(2)
  #pragma GCC optimize(3)
  using namespace std;
  const long long mod = 998244353;
  inline long long pw(long long x, long long y) {
    long long r = 1;
    while (y) {
      if (y & 1) r = r * x % mod;
      x = x * x % mod;
      y >>= 1;
    }
    return r;
  }
  template <class T1, class T2>
  inline const pair<T1, T2> operator+(const pair<T1, T2> &x,
                                      const pair<T1, T2> &y) {
    pair<T1, T2> z;
    z.first = x.first + y.first;
    z.second = x.second + y.second;
    return z;
  }
  template <class T1, class T2>
  inline const pair<T1, T2> operator-(const pair<T1, T2> &x,
                                      const pair<T1, T2> &y) {
    pair<T1, T2> z;
    z.first = x.first - y.first;
    z.second = x.second - y.second;
    return z;
  }
  int n, a[100010];
  vector<pair<int, int> > b, c;
  long long ans;
  inline pair<int, int> calc(int x, int y) {
    if (x <= y) return make_pair(x, 0);
    int z = (x - 1) / y + 1;
    return make_pair(x / z, z - 1);
  }
  void add(int x, int y) {
    if (!b.empty() && b.back().first == x) {
      b.back().second += y;
    } else {
      b.emplace_back(make_pair(x, y));
    }
  }
  void work() {
    cin >> n;
    ans = 0;
    for (register int i = 1; i <= n; i++) cin >> a[i];
    b.clear();
    c.clear();
    for (register int x = n; x >= 1; x--) {
      b.clear();
      pair<int, int> res;
      for (pair<int, int> i : c) {
        res = calc(a[x], i.first);
        add(res.first, i.second);
        {
          ans = ((ans + (long long)i.second * res.second % mod * x) % mod + mod) %
                mod;
        };
      }
      add(a[x], 1);
      c = b;
    }
    cout << ans << endl;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--) work();
    return 0;
  }
  