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;
  namespace Ruri {
  const int infi = 1e9;
  const long long infl = 4e18;
  }  // namespace Ruri
  using namespace Ruri;
  namespace Gadgets {
  inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
  inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
  }  // namespace Gadgets
  namespace Read {
  inline int ri() {
    int x;
    scanf("%d", &x);
    return x;
  }
  inline long long rl() {
    long long x;
    scanf("%lld", &x);
    return x;
  }
  inline double rd() {
    double x;
    scanf("%lf", &x);
    return x;
  }
  }  // namespace Read
  namespace DeBug {}
  using namespace Gadgets;
  using namespace Read;
  using namespace DeBug;
  const int MAX_N = 1e5 + 5;
  const int mod = 998244353;
  void Add(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  void Minus(int &x, int y) {
    x -= y;
    x += (x >> 31) & mod;
  };
  void Mul(int &x, int y) { x = 1ll * x * y % mod; }
  int n, a[MAX_N];
  int dp[2][MAX_N];
  vector<int> rec[2];
  int main() {
    memset(dp, -1, sizeof(dp));
    int T = ri();
    while (T--) {
      rec[0].clear(), rec[1].clear();
      n = ri();
      for (int i = 1, bbb = n; i <= bbb; ++i) a[i] = ri();
      int ans = 0;
      int now = 1, pre = 0;
      dp[pre][a[n]] = 1;
      rec[pre].push_back(a[n]);
      for (int i = n - 1, bbb = 1; i >= bbb; --i) {
        for (auto val : rec[pre]) {
          int k = (a[i] + val - 1) / val,
              res = 1ll * i * (k - 1) % mod * dp[pre][val] % mod;
          Add(ans, res);
          int now_val = a[i] / k;
          if (dp[now][now_val] == -1)
            rec[now].push_back(now_val), dp[now][now_val] = 0;
          Add(dp[now][now_val], dp[pre][val]);
          dp[pre][val] = -1;
        }
        if (dp[now][a[i]] == -1) rec[now].push_back(a[i]), dp[now][a[i]] = 0;
        Add(dp[now][a[i]], 1), rec[pre].clear();
        swap(pre, now);
      }
      for (auto val : rec[pre]) dp[pre][val] = -1;
      printf("%lld\n", pos_mod(ans, mod));
    }
    return 0;
  }
  