#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
const long long mod = 998244353;
inline long long pw(long long x, long long y) {
  long long r = 1;
  while (y) {
    if (y & 1) r = r * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return r;
}
template <class T1, class T2>
inline const pair<T1, T2> operator+(const pair<T1, T2> &x,
                                    const pair<T1, T2> &y) {
  pair<T1, T2> z;
  z.first = x.first + y.first;
  z.second = x.second + y.second;
  return z;
}
template <class T1, class T2>
inline const pair<T1, T2> operator-(const pair<T1, T2> &x,
                                    const pair<T1, T2> &y) {
  pair<T1, T2> z;
  z.first = x.first - y.first;
  z.second = x.second - y.second;
  return z;
}
int n, a[100010];
vector<pair<int, int> > b, c;
long long ans;
inline pair<int, int> calc(int x, int y) {
  if (x <= y) return make_pair(x, 0);
  int z = (x - 1) / y + 1;
  return make_pair(x / z, z - 1);
}
void add(int x, int y) {
  if (!b.empty() && b.back().first == x) {
    b.back().second += y;
  } else {
    b.emplace_back(make_pair(x, y));
  }
}
void work() {
  cin >> n;
  ans = 0;
  for (register int i = 1; i <= n; i++) cin >> a[i];
  b.clear();
  c.clear();
  for (register int x = n; x >= 1; x--) {
    b.clear();
    pair<int, int> res;
    for (pair<int, int> i : c) {
      res = calc(a[x], i.first);
      add(res.first, i.second);
      {
        ans = ((ans + (long long)i.second * res.second % mod * x) % mod + mod) %
              mod;
      };
    }
    add(a[x], 1);
    c = b;
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) work();
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
const long long mod = 998244353;
inline long long pw(long long x, long long y) {
  long long r = 1;
  while (y) {
    if (y & 1) r = r * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return r;
}
template <class T1, class T2>
inline const pair<T1, T2> operator+(const pair<T1, T2> &x,
                                    const pair<T1, T2> &y) {
  pair<T1, T2> z;
  z.first = x.first + y.first;
  z.second = x.second + y.second;
  return z;
}
template <class T1, class T2>
inline const pair<T1, T2> operator-(const pair<T1, T2> &x,
                                    const pair<T1, T2> &y) {
  pair<T1, T2> z;
  z.first = x.first - y.first;
  z.second = x.second - y.second;
  return z;
}
int n, a[100010];
vector<pair<int, int> > b, c;
long long ans;
inline pair<int, int> calc(int x, int y) {
  if (x <= y) return make_pair(x, 0);
  int z = (x - 1) / y + 1;
  return make_pair(x / z, z - 1);
}
void add(int x, int y) {
  if (!b.empty() && b.back().first == x) {
    b.back().second += y;
  } else {
    b.emplace_back(make_pair(x, y));
  }
}
void work() {
  cin >> n;
  ans = 0;
  for (register int i = 1; i <= n; i++) cin >> a[i];
  b.clear();
  c.clear();
  for (register int x = n; x >= 1; x--) {
    b.clear();
    pair<int, int> res;
    for (pair<int, int> i : c) {
      res = calc(a[x], i.first);
      add(res.first, i.second);
      {
        ans = ((ans + (long long)i.second * res.second % mod * x) % mod + mod) %
              mod;
      };
    }
    add(a[x], 1);
    c = b;
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) work();
  return 0;
}

```