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;
  void print_err() { cerr << "\n"; }
  template <class T, class... Arg>
  void print_err(T x, Arg &&...args) {
    cerr << x << " ";
    print_err(args...);
  }
  template <class T>
  void print_container(T &cont) {
    for (auto iter : cont) {
      cerr << iter << " ";
    }
    cerr << "\n";
  }
  int mod = 998244353;
  long long ceil(long long a, long long b) { return (a + b - 1) / b; }
  long long gcd(long long a, long long b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  vector<int> d4x{0, 0, 1, -1};
  vector<int> d4y{-1, 1, 0, 0};
  vector<int> d8x{0, 0, 1, -1, 1, 1, -1, -1};
  vector<int> d8y{1, -1, 0, 0, -1, 1, 1, -1};
  vector<long long> B(1e5 + 1, -1), C(1e5 + 1, -1);
  void solve(int test) {
    ;
    int n;
    cin >> n;
    vector<int> A(n);
    for (int &a : A) {
      cin >> a;
    }
    long long ans = 0;
    vector<long long> dp1, dp2;
    int j;
    long long val;
    for (int i = n - 1; i >= 0; --i) {
      for (auto iter : dp1) {
        j = iter;
        val = B[j];
        if (val < 0) {
          ;
        }
        if (val > 0) {
          if (j >= A[i]) {
            if (C[A[i]] == -1) {
              dp2.push_back(A[i]);
              C[A[i]] = val;
              C[A[i]] %= mod;
            } else {
              C[A[i]] += val;
              C[A[i]] %= mod;
            }
          } else {
            if (C[A[i] / ceil(A[i], j)] == -1) {
              dp2.push_back(A[i] / ceil(A[i], j));
              C[A[i] / ceil(A[i], j)] = val;
              ans += (i + 1) * (ceil(A[i], j) - 1) * val;
              ans %= mod;
              C[A[i] / ceil(A[i], j)] %= mod;
            } else {
              C[A[i] / ceil(A[i], j)] += val;
              ans += (i + 1) * (ceil(A[i], j) - 1) * val;
              ans %= mod;
              C[A[i] / ceil(A[i], j)] %= mod;
            }
          }
        }
        B[j] = -1;
      }
      dp1.clear();
      swap(dp1, dp2);
      swap(B, C);
      if (B[A[i]] == -1) {
        dp1.push_back(A[i]);
        B[A[i]] = 1;
      } else {
        B[A[i]]++;
        B[A[i]] %= mod;
      }
    }
    for (int j : dp1) {
      B[j] = -1;
    }
    cout << ans << endl;
  }
  int main() {
    int i = 1;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int test_cases;
    cin >> test_cases;
    for (i = 1; i <= test_cases; ++i) solve(i);
    return 0;
  }
  