#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 998244353;
int dp[2][100100], dp2[2][100100];
void clr(int pos, vector<int> &v) {
  for (auto x : v) dp[pos][x] = dp2[pos][x] = 0;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> val;
  int cur = 0, ans = 0;
  val.emplace_back(a[n - 1]);
  dp2[cur][a[n - 1]]++;
  for (int i = n - 2; i >= 0; i--) {
    cur ^= 1;
    vector<int> nval;
    for (auto x : val) {
      int dal = (a[i] + x - 1) / x;
      int cal = a[i] / dal;
      dp[cur][cal] =
          (dp[cur][cal] +
           ((dp[cur ^ 1][x] + 1LL * dp2[cur ^ 1][x] * (dal - 1) % mod) % mod) %
               mod) %
          mod;
      dp2[cur][cal] = (dp2[cur][cal] + dp2[cur ^ 1][x]) % mod;
      if (nval.size() == 0 || nval.back() != cal) nval.emplace_back(cal);
    }
    dp2[cur][a[i]] = (dp2[cur][a[i]] + 1) % mod;
    if (nval.size() == 0 || nval.back() != a[i]) nval.emplace_back(a[i]);
    for (auto x : nval) ans = (ans + dp[cur][x]) % mod;
    clr((cur ^ 1), val);
    val = nval;
  }
  clr(cur, val);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your task is to create 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;
using ll = long long;
const int mod = 998244353;
int dp[2][100100], dp2[2][100100];
void clr(int pos, vector<int> &v) {
  for (auto x : v) dp[pos][x] = dp2[pos][x] = 0;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> val;
  int cur = 0, ans = 0;
  val.emplace_back(a[n - 1]);
  dp2[cur][a[n - 1]]++;
  for (int i = n - 2; i >= 0; i--) {
    cur ^= 1;
    vector<int> nval;
    for (auto x : val) {
      int dal = (a[i] + x - 1) / x;
      int cal = a[i] / dal;
      dp[cur][cal] =
          (dp[cur][cal] +
           ((dp[cur ^ 1][x] + 1LL * dp2[cur ^ 1][x] * (dal - 1) % mod) % mod) %
               mod) %
          mod;
      dp2[cur][cal] = (dp2[cur][cal] + dp2[cur ^ 1][x]) % mod;
      if (nval.size() == 0 || nval.back() != cal) nval.emplace_back(cal);
    }
    dp2[cur][a[i]] = (dp2[cur][a[i]] + 1) % mod;
    if (nval.size() == 0 || nval.back() != a[i]) nval.emplace_back(a[i]);
    for (auto x : nval) ans = (ans + dp[cur][x]) % mod;
    clr((cur ^ 1), val);
    val = nval;
  }
  clr(cur, val);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```