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;
  const int N = 1e5 + 10;
  const long long mod = 998244353;
  inline long long gcd(long long a, long long b) {
    return !b ? a : gcd(b, a % b);
  }
  inline long long q_pow(long long a, long long x = mod - 2) {
    long long ans = 1, tmp = a;
    while (x) {
      if (x & 1) (ans *= tmp) %= mod;
      (tmp *= tmp) %= mod;
      x >>= 1;
    }
    return ans;
  }
  template <typename T>
  inline void re(T &N) {
    int f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9')
      if (c == '-') f = -1;
    N = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
    N *= f;
  }
  template <class T, class... T_>
  inline void re(T &x, T_ &...y) {
    re(x), re(y...);
  }
  int m, n, t = 1, st, en;
  int a[N], b[N], c[N], d[N];
  struct SegT {
    long long e[N << 2], tag[N << 2];
    inline void push_up(int p) {
      e[p] = (e[((p) << 1)] + e[((p) << 1 | 1)]) % mod;
    }
    void build(int p, int l, int r) {
      e[p] = tag[p] = 0;
      if (l == r) return;
      int mid = (l + r) >> 1;
      build(((p) << 1), l, mid);
      build(((p) << 1 | 1), mid + 1, r);
    }
    inline void push_down(int p, int l, int r) {
      if (tag[p]) {
        int mid = (l + r) >> 1;
        (e[((p) << 1)] += 1ll * (mid - l + 1) * tag[p] % mod) %= mod;
        (tag[((p) << 1)] += tag[p]) %= mod;
        (e[((p) << 1 | 1)] += 1ll * (r - mid) * tag[p] % mod) %= mod;
        (tag[((p) << 1 | 1)] += tag[p]) %= mod;
        tag[p] = 0;
      }
    }
    void update(int p, int l, int r, int L, int R, long long k) {
      if (L <= l && r <= R)
        return (e[p] += 1ll * (r - l + 1) * k % mod) %= mod, (tag[p] += k) %= mod,
               void();
      int mid = (l + r) >> 1;
      push_down(p, l, r);
      if (L <= mid) update(((p) << 1), l, mid, L, R, k);
      if (mid < R) update(((p) << 1 | 1), mid + 1, r, L, R, k);
      push_up(p);
    }
    long long query(int p, int l, int r, int L, int R) {
      long long ans = 0;
      if (L <= l && r <= R) return e[p];
      int mid = (l + r) >> 1;
      push_down(p, l, r);
      if (L <= mid) (ans += query(((p) << 1), l, mid, L, R)) %= mod;
      if (mid < R) (ans += query(((p) << 1 | 1), mid + 1, r, L, R)) %= mod;
      return ans;
    }
  } tr;
  int main() {
    re(t);
    while (t--) {
      re(n);
      tr.build(1, 1, n);
      for (int i = 1; i <= n; i++) re(a[i]), b[i] = c[i] = d[i] = 0;
      long long ans = 0, now = 0;
      for (int i = 1; i <= n; i++) {
        auto calc = [&](int x, int y) { return x / ((x + y - 1) / y); };
        auto times = [&](int x, int y) { return (x + y - 1) / y - 1; };
        c[i] = a[i];
        d[i] = 0;
        for (int j = i - 1; j >= 1; j--) {
          now -= 1ll * d[j] * j;
          d[j] = times(a[j], c[j + 1]);
          now += 1ll * d[j] * j;
          if (calc(a[j], c[j + 1]) == c[j]) break;
          c[j] = calc(a[j], c[j + 1]);
        }
        (ans += now) %= mod;
      }
      printf("%lld\n", ans);
    }
  }
  