#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = ' ';
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') f = -1;
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch - 48), ch = getchar();
  return x * f;
}
const int mod = 998244353;
int t, n, a[100005];
vector<pair<long long, long long> > g[2];
int main() {
  cin >> t;
  while (t--) {
    n = read();
    long long ans = 0;
    for (int i = 1; i <= n; i++) a[i] = read();
    g[n % 2].clear();
    g[n % 2].push_back({a[n], 1});
    for (int i = n; i > 1; i--) {
      int las = (i % 2), now = (las ^ 1);
      g[now].clear();
      int las1 = 0;
      for (auto tmp : g[las]) {
        ans += ((a[i - 1] - 1) / tmp.first) * (i - 1) % mod * tmp.second % mod;
        int tmp1 = a[i - 1] / ((a[i - 1] - 1) / tmp.first + 1);
        if (tmp1 == las1)
          g[now][g[now].size() - 1].second += tmp.second % mod;
        else
          g[now].push_back({tmp1, tmp.second}), las1 = tmp1;
      }
      if (a[i - 1] == las1)
        g[now][g[now].size() - 1].second++;
      else
        g[now].push_back({a[i - 1], 1ll});
      g[las] = g[now];
    }
    cout << ans % mod << endl;
  }
}

### Prompt

Create a solution in cpp for 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;
long long read() {
  long long x = 0, f = 1;
  char ch = ' ';
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') f = -1;
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch - 48), ch = getchar();
  return x * f;
}
const int mod = 998244353;
int t, n, a[100005];
vector<pair<long long, long long> > g[2];
int main() {
  cin >> t;
  while (t--) {
    n = read();
    long long ans = 0;
    for (int i = 1; i <= n; i++) a[i] = read();
    g[n % 2].clear();
    g[n % 2].push_back({a[n], 1});
    for (int i = n; i > 1; i--) {
      int las = (i % 2), now = (las ^ 1);
      g[now].clear();
      int las1 = 0;
      for (auto tmp : g[las]) {
        ans += ((a[i - 1] - 1) / tmp.first) * (i - 1) % mod * tmp.second % mod;
        int tmp1 = a[i - 1] / ((a[i - 1] - 1) / tmp.first + 1);
        if (tmp1 == las1)
          g[now][g[now].size() - 1].second += tmp.second % mod;
        else
          g[now].push_back({tmp1, tmp.second}), las1 = tmp1;
      }
      if (a[i - 1] == las1)
        g[now][g[now].size() - 1].second++;
      else
        g[now].push_back({a[i - 1], 1ll});
      g[las] = g[now];
    }
    cout << ans % mod << endl;
  }
}

```