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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long R = 998244353, R1 = 19491001, R2 = 236, NUMTESTCASE;
const long long NN = 10 + 1e5, Inverse2 = 500000004;
const double pi = acos(-1.0);
int di[8] = {1, 0, -1, 0, 1, -1, 1, -1}, dj[8] = {0, 1, 0, -1, 1, -1, -1, 1};
int n, Arr[NN], Total;
map<int, pair<int, int>> Memo[NN];
pair<int, int> Dp(int Pos, int l, int Next) {
  if (Pos < l) return {0, Next};
  if (Memo[Pos].count(Next)) return Memo[Pos][Next];
  Dp(Pos - 1, l, INT_MAX);
  pair<int, int> res;
  if (Arr[Pos] <= Next)
    res = {1 + Dp(Pos - 1, l, Arr[Pos]).first, Dp(Pos - 1, l, Arr[Pos]).second};
  else {
    int Number = (Arr[Pos] + Next - 1) / Next;
    int p = Arr[Pos] / Number;
    res = {Number + Dp(Pos - 1, l, p).first, Dp(Pos - 1, l, p).second};
  }
  if (Next == INT_MAX) Total += res.first;
  return Memo[Pos][Next] = res;
}
int main() {
  for (cin >> NUMTESTCASE; NUMTESTCASE; NUMTESTCASE--) {
    scanf("%d", &n);
    for (int i = (1); i <= (n); ++i) scanf("%d", Arr + i);
    int Ans = 0;
    for (long long i = 1; i <= n; i++)
      Ans = (Ans + (n - i + 2) * (n - i + 1) / 2) % R;
    Ans = (R - Ans) % R;
    map<int, pair<int, int>> dp;
    for (int i = (n); i >= (1); --i) {
      map<int, pair<int, int>> dp2;
      dp2[Arr[i]] = {1, 1};
      int c = 1;
      for (auto p : dp) {
        int t = 0, Nxt = Arr[i], Cnt = p.second.second;
        if (Arr[i] <= p.first)
          t = 1;
        else
          t = (Arr[i] + p.first - 1) / p.first,
          Nxt = Arr[i] / ((Arr[i] + p.first - 1) / p.first);
        dp2[Nxt] = {dp2[Nxt].first + t * Cnt + p.second.first,
                    dp2[Nxt].second + Cnt};
        c += Cnt;
      }
      assert(c == n - i + 1);
      dp = dp2;
      for (auto p : dp) Ans = (Ans + p.second.first) % R;
    }
    printf("%d\n", Ans);
  }
  return 0;
}
