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 MOD = 998244353;
  const int inf = 0x3f3f3f3f;
  const int N = 110000;
  template <typename T>
  inline bool scan_d(T& ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != '-' && (c < '0' || c > '9')) {
      if ((c = getchar()) == EOF) return 0;
    }
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
  }
  template <typename T>
  void print(T x) {
    static char s[33], *s1;
    s1 = s;
    if (!x) *s1++ = '0';
    if (x < 0) putchar('-'), x = -x;
    while (x) *s1++ = (x % 10 + '0'), x /= 10;
    while (s1-- != s) putchar(*s1);
  }
  template <typename T>
  void println(T x) {
    print(x);
    putchar('\n');
  }
  int norm(int x) {
    if (x < 0) {
      x += MOD;
    }
    if (x >= MOD) {
      x -= MOD;
    }
    return x;
  }
  template <class T>
  T power(T a, int b) {
    T res = 1;
    for (; b; b >>= 1, a *= a) {
      if (b & 1) {
        res *= a;
      }
    }
    return res;
  }
  struct mint {
    int x;
    mint(int x = 0) : x(norm(x)) {}
    int val() const { return x; }
    mint operator-() const { return mint(norm(MOD - x)); }
    mint inv() const { return power(*this, MOD - 2); }
    mint& operator*=(const mint& rv) {
      x = (long long)x * rv.x % MOD;
      return *this;
    }
    mint& operator+=(const mint& rv) {
      x = norm(x + rv.x);
      return *this;
    }
    mint& operator-=(const mint& rv) {
      x = norm(x - rv.x);
      return *this;
    }
    mint& operator/=(const mint& rv) { return *this *= rv.inv(); }
    friend mint operator*(const mint& lv, const mint& rv) {
      mint res = lv;
      res *= rv;
      return res;
    }
    friend mint operator+(const mint& lv, const mint& rv) {
      mint res = lv;
      res += rv;
      return res;
    }
    friend mint operator-(const mint& lv, const mint& rv) {
      mint res = lv;
      res -= rv;
      return res;
    }
    friend mint operator/(const mint& lv, const mint& rv) {
      mint res = lv;
      res /= rv;
      return res;
    }
  };
  struct TPoint {
    int v, x, y;
  };
  int n;
  int a[N];
  vector<TPoint> b[2];
  vector<long long> dp[2];
  void fmain() {
    scan_d(n);
    for (int i = (1); i < (n + 1); ++i) scan_d(a[i]);
    mint res = 0;
    int las = 0, cur = 0;
    for (int i = (1); i < (n + 1); ++i) {
      las = cur, cur ^= 1;
      b[cur].clear();
      for (int j = (1); j < (a[i] + 1); ++j) {
        if (j * j > a[i]) break;
        int v = j, x = a[i] / j, y = a[i] / j;
        if (a[i] % j != 0) y++;
        b[cur].push_back({v, x, y});
      }
      vector<TPoint> bb;
      for (int j = (0); j < (a[i] + 1); ++j) {
        if (j * j > a[i]) break;
        int v = (a[i] + j) / (j + 1), x = (j + 1), y = (j + 1);
        if (a[i] % (j + 1) != 0) x--;
        bb.push_back({v, x, y});
      }
      for (int j = (bb.size()) - 1; j >= (0); --j) b[cur].push_back(bb[j]);
      dp[cur].resize(b[cur].size());
      if (i == 1) {
        for (int j = (0); j < (dp[cur].size()); ++j) dp[cur][j] = b[cur][j].v - 1;
      } else {
        int k = 0;
        for (int j = (0); j < (dp[cur].size()); ++j) {
          while (b[las][k].y > b[cur][j].x) k++;
          dp[cur][j] = dp[las][k] + (long long)i * (b[cur][j].v - 1);
        }
      }
      res += (int)(dp[cur][0] % MOD);
    }
    println(res.val());
  }
  int main() {
    int T;
    scan_d(T);
    for (int t = (0); t < (T); ++t) {
      fmain();
    }
    return 0;
  }
  