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 debug_out() { cerr << "\n"; }
  template <typename H, typename... T>
  void debug_out(H h, T... t) {
    cerr << " " << (h);
    debug_out(t...);
  }
  const int MOD = 998244353;
  long long mul(long long a, long long b) { return (a * b) % MOD; }
  long long add(long long a, long long b) { return (a + b) % MOD; }
  const int N = 1e5 + 2;
  const int S = 702;
  int b[2][S];
  int siz[N];
  int dp[2][S];
  int a[N];
  void init(int i) {
    auto &bb = b[i % 2];
    int x = a[i];
    siz[i] = 0;
    for (int k = 1; k <= min(S, x); ++k) {
      if (siz[i] and x / k == bb[siz[i] - 1]) break;
      bb[siz[i]++] = x / k;
    }
    for (int val = 1; val <= min(S, x); ++val) {
      if (x / (x / val) != val) break;
      bb[siz[i]++] = val;
    }
    sort(bb, bb + siz[i]);
    siz[i] = unique(bb, bb + siz[i]) - bb;
  }
  void solve() {
    int n, res;
    cin >> n;
    fill(siz, siz + n + 1, 0);
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    fill(dp[0], dp[0] + S, 0);
    fill(dp[1], dp[1] + S, 0);
    res = 0;
    init(n);
    dp[n % 2][siz[n]] = 1;
    int sz, x, y, ad, k, pos, s, idx;
    for (int i = n - 1; i >= 1; --i) {
      init(i);
      auto &b1 = b[(i + 1) % 2];
      sz = siz[i + 1];
      auto &bb = b[i % 2];
      s = siz[i];
      auto &cur = dp[i % 2];
      fill(cur, cur + S, 0);
      cur[s] = 1;
      auto &nxt = dp[(i + 1) % 2];
      idx = 0;
      for (int j = 0; j < sz; ++j) {
        ad = nxt[j + 1];
        if (ad == 0) continue;
        y = b1[j];
        if (a[i] <= y) {
          cur[s] = add(cur[s], ad);
          continue;
        }
        k = (a[i] + y - 1) / y;
        x = a[i] / k;
        while (idx < s and bb[idx] < x) ++idx;
        assert(idx < s and bb[idx] == x);
        pos = idx + 1;
        cur[pos] = add(cur[pos], ad);
        res = add(res, mul(k - 1, mul(ad, i)));
      }
    }
    cout << res << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    for (int i = 1; i <= t; ++i) {
      solve();
    }
    return 0;
  }
  