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 long long maxn = 1e5 + 5, m = 998244353;
  long long bin_pow(long long a, long long b) {
    long long x = 1;
    while (b) {
      if (b & 1) x *= a;
      a *= a;
      b >>= 1;
    }
  }
  long long readint() {
    long long x = 0, f = 0;
    char c = getchar();
    while (!isdigit(c)) c = getchar(), f |= c == '-';
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    return f ? -x : x;
  }
  string readstring() {
    string s = "";
    char c = getchar();
    while (!isalpha(c)) c = getchar();
    while (isalpha(c)) s += c, c = getchar();
  }
  void print(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  void println(long long x) {
    print(x);
    putchar('\n');
  }
  void printsp(long long x) {
    print(x);
    putchar(' ');
  }
  void printyn(bool i) {
    if (i) {
      putchar('Y');
      putchar('E');
      putchar('S');
    } else {
      putchar('N');
      putchar('O');
    }
    putchar('\n');
  };
  struct dsu {
    long long p[maxn], r[maxn];
    long long n;
    dsu(long long n) {
      this->n = n;
      for (long long i = 0; i < n; ++i) p[i] = 1, r[i] = 0;
    }
    long long find(long long u) { return p[u] == u ? u : p[u] = find(p[u]); }
    bool unite(long long u, long long v) {
      u = find(u);
      v = find(v);
      if (r[u] < r[v]) swap(u, v);
      p[v] = u;
      r[u] += r[v];
    }
  };
  struct fw {
    long long t[maxn];
    long long n;
    fw(long long n) { this->n = n; }
    void upd(long long i, long long d) {
      for (; i < n; i |= i + 1) t[i] += d;
    }
    long long sum(long long i) {
      long long x = 0;
      for (; i >= 0; i &= i + 1, --i) x += t[i];
      return x;
    }
  };
  long long max(long long a, long long b) { return a > b ? a : b; }
  long long min(long long a, long long b) { return a > b ? b : a; }
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long len(long long x) {
    if (x > 0) return len(x / 10) + 1;
  }
  long long len2(long long x) {
    if (x == 1)
      return 1;
    else
      return len2(x - 1) * 10;
  }
  long long a[maxn], dp[2][maxn];
  vector<long long> b[2];
  signed main() {
    long long t = readint();
    while (t--) {
      long long n = readint(), ans = 0;
      for (long long i = 1; i <= n; ++i) a[i] = readint();
      for (long long i = n; i; --i) {
        long long j = i & 1, l = a[i];
        b[j].push_back(a[i]);
        dp[j][a[i]] = 1;
        for (long long x : b[j ^ 1]) {
          long long y = dp[j ^ 1][x], s = (a[i] + x - 1) / x, st = a[i] / s;
          ans += (s - 1) * y * i;
          dp[j][st] += y;
          if (l != st) b[j].push_back(st), l = st;
        }
        for (long long x : b[j ^ 1]) dp[j ^ 1][x] = 0;
        b[j ^ 1].clear();
        ans %= m;
      }
      println(ans);
      for (long long i = 0; i < 2; ++i)
        for (long long x : b[i]) dp[i][x] = 0;
      b[0].clear();
      b[1].clear();
    }
  }
  