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;
  template <typename T>
  inline void ckmax(T &x, T y) {
    x = x < y ? y : x;
    return;
  }
  template <typename T>
  inline void ckmin(T &x, T y) {
    x = x < y ? x : y;
    return;
  }
  namespace FastIO {
  inline long long read() {
    long long res = 0, f = 1;
    char k;
    while (!isdigit(k = getchar()))
      if (k == '-') f = -1;
    while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
    return res * f;
  }
  char OPUT[100];
  inline void print(long long x) {
    if (!x) return putchar('0'), putchar('\n'), void();
    if (x < 0) putchar('-'), x = -x;
    long long cnt = 0;
    while (x) OPUT[++cnt] = x % 10, x /= 10;
    for (long long i = cnt; i >= 1; --i) putchar(OPUT[i] + '0');
    putchar('\n');
    return;
  }
  }  // namespace FastIO
  using namespace FastIO;
  const long long mod = 998244353;
  inline long long add(long long x, long long y, long long Mod = mod) {
    return x + y >= Mod ? x + y - Mod : x + y;
  }
  inline long long del(long long x, long long y, long long Mod = mod) {
    return x - y < 0 ? x - y + Mod : x - y;
  }
  inline long long mul(long long x, long long y, long long Mod = mod) {
    return x * y - x * y / Mod * Mod;
  }
  inline void ckadd(long long &x, long long y, long long Mod = mod) {
    x = x + y >= Mod ? x + y - Mod : x + y;
  }
  inline void ckdel(long long &x, long long y, long long Mod = mod) {
    x = x - y < 0 ? x - y + Mod : x - y;
  }
  inline void ckmul(long long &x, long long y, long long Mod = mod) {
    x = x * y - x * y / Mod * Mod;
  }
  inline long long ksm(long long x, long long y, long long Mod = mod) {
    long long res = 1;
    for (; y; y >>= 1, ckmul(x, x, Mod))
      if (y & 1) ckmul(res, x, Mod);
    return res;
  }
  inline void approx(long long val, long long Mod = mod, long long lim = 1e5) {
    long long x = val, y = Mod, a = 1, b = 0;
    while (x > lim) {
      swap(x, y);
      swap(a, b);
      a -= x / y * b;
      x %= y;
    }
    cout << x << "/" << a << endl;
    return;
  }
  const long long N = 1e5 + 10;
  long long dp[2][N], vis[2][N], a[N], n;
  vector<long long> app[2];
  inline void insert(long long cur, long long val) {
    if (!vis[cur][val]) vis[cur][val] = 1, app[cur].push_back(val);
    return;
  }
  signed main() {
    long long T = read();
    while (T--) {
      n = read();
      for (long long i = 1; i <= n; ++i) a[i] = read();
      long long cur = 0, ans = 0;
      for (long long i = n; i >= 1; --i, cur ^= 1) {
        ckadd(dp[cur][a[i]], 1);
        insert(cur, a[i]);
        for (auto t : app[cur ^ 1]) {
          long long k = (a[i] + t - 1) / t;
          ckadd(dp[cur][a[i] / k], dp[cur ^ 1][t]);
          insert(cur, a[i] / k);
          ckadd(ans, mul(i, mul(dp[cur ^ 1][t], k - 1)));
          dp[cur ^ 1][t] = vis[cur ^ 1][t] = 0;
        }
        app[cur ^ 1].clear();
      }
      cur ^= 1;
      for (auto t : app[cur]) vis[cur][t] = dp[cur][t] = 0;
      app[cur].clear();
      print(ans);
    }
    return 0;
  }
  