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;
  using namespace std;
  inline int read() {
    char ch;
    bool sign = 0;
    int res = 0;
    while (!isdigit(ch = getchar()))
      if (ch == '-') sign = 1;
    while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
    if (sign)
      return -res;
    else
      return res;
  }
  const int maxn = 100000 + 10;
  const int mod = 998244353;
  const int inf = 1e9;
  const int maxm = 640;
  int a[maxn], n;
  vector<int> val;
  inline int add(int x, int y) {
    x += y;
    if (x >= mod)
      x -= mod;
    else if (x < 0)
      x += mod;
    return x;
  }
  inline int mul(int x, int y) { return 1LL * x * y % mod; }
  int dp[2][maxm];
  inline void clear(int t) {
    for (int i = (0); i < (maxm); ++i) {
      dp[t][i] = 0;
    }
  }
  inline int solve() {
    int t = 0;
    clear(t);
    int ans = 0;
    val.clear();
    for (int l = 1; l <= a[1];) {
      val.push_back(a[1] / l);
      l = a[1] / val.back() + 1;
    }
    for (int i = (2); i <= (n); ++i) {
      clear(t ^ 1);
      int now = 0;
      vector<int> nowval;
      int l = 1;
      int c = 0;
      for (; l <= a[i];) {
        nowval.push_back(a[i] / l);
        int j = nowval[c];
        l = a[i] / j + 1;
        int tmp = a[i - 1] / ((a[i - 1] + j - 1) / j);
        for (; val[now] != tmp && now >= 0;) {
          ++now;
        }
        assert(now >= 0);
        assert(val[now] == tmp);
        int v = a[i - 1] / j;
        if (a[i - 1] % j == 0) v--;
        dp[t ^ 1][c] = add(dp[t][now], mul(i - 1, v));
        c++;
      }
      ans = add(ans, dp[t ^ 1][0]);
      val = nowval;
      t ^= 1;
    }
    return ans;
  }
  int main(void) {
    int T;
    scanf("%d", &(T));
    for (; T--;) {
      scanf("%d", &(n));
      for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
      printf("%d\n", solve());
    }
    return 0;
  }
  