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.
#include <bits/stdc++.h>
using namespace std;
namespace FullPre {
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
template <typename T, typename... args>
inline void read(T &x, args &...Args) {
  read(x), read(Args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void print_char(T x, char ch) {
  write(x), putchar(ch);
}
template <typename T>
inline void char_print(char ch, T x) {
  putchar(ch), write(x);
}
template <typename T>
inline void print_space(T x) {
  print_char(x, ' ');
}
template <typename T>
inline void print_space_mul(T x) {
  print_space(x);
}
template <typename T, typename... args>
inline void print_space_mul(T x, args... Args) {
  print_space(x), print_space_mul(Args...);
}
template <typename T>
inline void space_print(T x) {
  char_print(' ', x);
}
template <typename T>
inline void space_print_mul(T x) {
  space_print(x);
}
template <typename T, typename... args>
inline void space_print_mul(T x, args... Args) {
  space_print(x), space_print_mul(Args...);
}
template <typename T>
inline void println(T x) {
  print_char(x, '\n');
}
template <typename T>
inline void println_mul(T x) {
  println(x);
}
template <typename T, typename... args>
inline void println_mul(T x, args... Args) {
  println(x), println_mul(Args...);
}
template <typename T>
inline void lnprint(T x) {
  char_print('\n', x);
}
template <typename T>
inline void lnprint_mul(T x) {
  lnprint(x);
}
template <typename T, typename... args>
inline void lnprint_mul(T x, args... Args) {
  lnprint(x), lnprint_mul(Args...);
}
inline void filein(string s) { freopen((s + ".in").c_str(), "r", stdin); }
inline void fileout(string s) { freopen((s + ".out").c_str(), "w", stdout); }
inline void file(string s) { filein(s), fileout(s); }
template <typename T>
inline T mymin(T a, T b) {
  return a < b ? a : b;
}
template <typename T, typename... args>
inline T mymin(T a, T b, args... Args) {
  return mymin(mymin(a, b), Args...);
}
template <typename T>
inline T mymax(T a, T b) {
  return a > b ? a : b;
}
template <typename T, typename... args>
inline T mymax(T a, T b, args... Args) {
  return mymax(mymax(a, b), Args...);
}
template <typename T>
inline T myabs(T a) {
  return a < 0 ? -a : a;
}
template <typename T>
inline T mysq(T a) {
  return a * a;
}
template <typename T>
inline T mygcd(T a, T b) {
  return !b ? a : mygcd(b, a % b);
}
template <typename T, typename... args>
inline T mygcd(T a, T b, args... Args) {
  return mygcd(mygcd(a, b), Args...);
}
template <typename T>
inline T mylcm(T a, T b) {
  return a / mygcd(a, b) * b;
}
template <typename T, typename... args>
inline T mylcm(T a, T b, args... Args) {
  return mylcm(mylcm(a, b), Args...);
}
template <typename T>
inline T myswap(T &a, T &b) {
  a ^= b ^= a ^= b;
}
}  // namespace FullPre
using namespace FullPre;
namespace Solution {
const int N = 1e5 + 7, mod = 998244353;
int n, a[N], f[N], sum[N], t[N];
inline void Main() {
  int Test;
  read(Test);
  while (Test--) {
    read(n);
    for (int(i) = (1); (i) <= (n); ++(i)) read(a[i]);
    int cur = a[n], ans = 0;
    for (int(i) = (n - 1); (i) >= (1); --(i)) {
      ans = (ans + ((a[i] - 1) / cur) * i) % mod;
      sum[i] = (a[i] - 1) / cur * i, cur = a[i] / ((a[i] - 1) / cur + 1),
      t[i] = cur;
    }
    for (int(i) = (1); (i) <= (n - 1); ++(i))
      sum[i] = (sum[i] + sum[i - 1]) % mod;
    for (int(i) = (n - 1); (i) >= (1); --(i)) {
      int cur = a[i], j = i - 1;
      for (; j >= 1; --j) {
        sum[j] = (a[j] - 1) / cur * j, cur = a[j] / ((a[j] - 1) / cur + 1);
        if (cur == t[j]) break;
        t[j] = cur;
      }
      for (int(k) = (j); (k) <= (i - 1); ++(k))
        if (k) sum[k] = (sum[k] + sum[k - 1]) % mod;
      ans = (ans + sum[i - 1]) % mod;
    }
    println(ans);
  }
  return;
}
}  // namespace Solution
int main() {
  Solution ::Main();
  return 0;
}
