#include <bits/stdc++.h>
using namespace std;
template <typename T>
void Print(T v) {
  for (auto i : v) cout << i << (" ");
  cout << '\n';
}
int INF = 1000000007;
long long LINF = 1000000000000000000;
long long mod1 = 998244353;
long long dp[2][100001];
long long ceil(int a, int b) {
  long long res = a / b;
  if (a % b) res++;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> vec(n);
  for (int i = 0; i < n; i++) cin >> vec[i];
  long long res = 0;
  dp[1][vec[n - 1]] = 1;
  vector<vector<long long>> iter(2);
  iter[1].push_back(vec[n - 1]);
  for (int i = n - 2; i >= 0; i--) {
    iter[0].push_back(vec[i]);
    dp[0][vec[i]] = 1;
    while (not iter[1].empty()) {
      long long pod = ceil(vec[i], iter[1].back());
      if (dp[0][vec[i] / pod] == 0) iter[0].push_back(vec[i] / pod);
      dp[0][vec[i] / pod] += dp[1][iter[1].back()];
      res += (i + 1) * dp[1][iter[1].back()] * (pod - 1);
      res %= mod1;
      dp[1][iter[1].back()] = 0;
      iter[1].pop_back();
    }
    for (auto j : iter[0]) swap(dp[1][j], dp[0][j]);
    swap(iter[0], iter[1]);
  }
  while (not iter[1].empty()) {
    dp[1][iter[1].back()] = 0;
    iter[1].pop_back();
  }
  cout << (res) << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
template <typename T>
void Print(T v) {
  for (auto i : v) cout << i << (" ");
  cout << '\n';
}
int INF = 1000000007;
long long LINF = 1000000000000000000;
long long mod1 = 998244353;
long long dp[2][100001];
long long ceil(int a, int b) {
  long long res = a / b;
  if (a % b) res++;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> vec(n);
  for (int i = 0; i < n; i++) cin >> vec[i];
  long long res = 0;
  dp[1][vec[n - 1]] = 1;
  vector<vector<long long>> iter(2);
  iter[1].push_back(vec[n - 1]);
  for (int i = n - 2; i >= 0; i--) {
    iter[0].push_back(vec[i]);
    dp[0][vec[i]] = 1;
    while (not iter[1].empty()) {
      long long pod = ceil(vec[i], iter[1].back());
      if (dp[0][vec[i] / pod] == 0) iter[0].push_back(vec[i] / pod);
      dp[0][vec[i] / pod] += dp[1][iter[1].back()];
      res += (i + 1) * dp[1][iter[1].back()] * (pod - 1);
      res %= mod1;
      dp[1][iter[1].back()] = 0;
      iter[1].pop_back();
    }
    for (auto j : iter[0]) swap(dp[1][j], dp[0][j]);
    swap(iter[0], iter[1]);
  }
  while (not iter[1].empty()) {
    dp[1][iter[1].back()] = 0;
    iter[1].pop_back();
  }
  cout << (res) << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

```