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 optimize(3)
  using namespace std;
  const int maxn = 1e5 + 3, maxm = 1e5 + 11, inf_int = 0x3f3f3f3f;
  const long long inf_ll = 0x3f3f3f3f3f3f3f, mod = 998244353;
  const double eps = 1e-6;
  const long long inv2 = (mod + 1) / 2;
  struct custom_hash {
    static size_t splitmix64(size_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(size_t x) const {
      static const size_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  int n, a[maxn], pre, cur;
  long long ans, en[maxn][303], lim, f[maxn];
  inline long long DP(long long x, long long v) {
    long long res = 0;
    for (int i = x; i; i--)
      if (v < lim) {
        res = (res + en[i][v]) % mod;
        break;
      } else if (v >= a[i] && i < x) {
        res = (res + f[i]) % mod;
        break;
      } else {
        int t0 = (a[i] + v - 1) / v;
        res = (res + 1ll * i * (t0 - 1) % mod) % mod;
        v = a[i] / t0;
      }
    return f[x] = res;
  }
  inline long long dfs(int x, int v) {
    if (x == 0) return 0;
    if (en[x][v] != -1) return en[x][v];
    int t0 = (a[x] + v - 1) / v;
    en[x][v] = (dfs(x - 1, a[x] / t0) + 1ll * x * (t0 - 1) % mod) % mod;
    return en[x][v];
  }
  inline void solve() {
    cin >> n, ans = 0, cur = 1, pre = 0;
    lim = 303;
    for (int i = 1; i <= n; i++) {
      cin >> a[i], f[i] = 0;
      for (int j = 1; j < lim; j++) en[i][j] = -1;
    }
    for (int i = 1; i < n; i++)
      for (int j = 1; j < lim; j++) dfs(i, j);
    for (int i = 1; i < n; i++) {
      ans = (ans + DP(i, a[i + 1])) % mod;
    }
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int TEST = 1;
    cin >> TEST;
    while (TEST--) solve();
  }
  