#include <bits/stdc++.h>
using namespace std;
const int64_t inf = 1e18;
const int64_t mod = 998244353;
int32_t main() {
  int64_t q;
  cin >> q;
  while (q--) {
    int64_t n;
    cin >> n;
    vector<int64_t> arr(n);
    for (int64_t& e : arr) {
      cin >> e;
    }
    vector<pair<int64_t, int64_t>> lst{{inf, 1}};
    int64_t ans = 0;
    for (int64_t i = n - 1; i >= 0; --i) {
      vector<pair<int64_t, int64_t>> new_lst;
      for (auto& e : lst) {
        int64_t d = (arr[i] + e.first - 1) / e.first;
        ans += e.second * (i + 1) * (d - 1) % mod;
        ans %= mod;
        new_lst.push_back({arr[i] / d, e.second});
      }
      lst.clear();
      lst.push_back(new_lst[0]);
      for (int64_t i = 1; i < new_lst.size(); ++i) {
        if (new_lst[i].first == lst.back().first) {
          lst.back().second += new_lst[i].second;
        } else {
          lst.push_back(new_lst[i]);
        }
      }
      lst.push_back({inf, 1});
    }
    cout << ans << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int64_t inf = 1e18;
const int64_t mod = 998244353;
int32_t main() {
  int64_t q;
  cin >> q;
  while (q--) {
    int64_t n;
    cin >> n;
    vector<int64_t> arr(n);
    for (int64_t& e : arr) {
      cin >> e;
    }
    vector<pair<int64_t, int64_t>> lst{{inf, 1}};
    int64_t ans = 0;
    for (int64_t i = n - 1; i >= 0; --i) {
      vector<pair<int64_t, int64_t>> new_lst;
      for (auto& e : lst) {
        int64_t d = (arr[i] + e.first - 1) / e.first;
        ans += e.second * (i + 1) * (d - 1) % mod;
        ans %= mod;
        new_lst.push_back({arr[i] / d, e.second});
      }
      lst.clear();
      lst.push_back(new_lst[0]);
      for (int64_t i = 1; i < new_lst.size(); ++i) {
        if (new_lst[i].first == lst.back().first) {
          lst.back().second += new_lst[i].second;
        } else {
          lst.push_back(new_lst[i]);
        }
      }
      lst.push_back({inf, 1});
    }
    cout << ans << endl;
  }
}

```