#include <bits/stdc++.h>
using namespace std;
long long I() {
  char c = getchar();
  long long x = 0;
  long long f = 1;
  while (c < '0' or c > '9') f = (c == '-') ? -1 : 1, c = getchar();
  while (c >= '0' and c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return ((f == 1) ? x : -x);
}
long long n, a[100005];
long long f[2][100005];
vector<long long> n0[2];
inline void ad(long long& x, long long y) { x = (x + y) % 998244353; }
void flandre() {
  n = I();
  for (long long i = 1; i <= n; ++i) a[i] = I();
  long long ans = 0;
  long long pre = 0, cur = 1;
  n0[0].clear();
  n0[1].clear();
  for (long long i = n; i >= 1; --i) {
    f[cur][a[i]] = 1;
    n0[cur].push_back(a[i]);
    long long lasp = -1;
    for (auto x : n0[pre]) {
      long long p = a[i] / ((a[i] + x - 1) / x);
      ad(f[cur][p], f[pre][x]);
      if (p != lasp and p != a[i]) n0[cur].push_back(p), lasp = p;
      ad(ans, f[pre][x] * i % 998244353 * ((a[i] - 1) / x) % 998244353);
    }
    for (auto x : n0[pre]) f[pre][x] = 0;
    n0[pre].clear();
    swap(pre, cur);
  }
  printf("%lld\n", ans);
  for (auto x : n0[0]) f[0][x] = 0;
  for (auto x : n0[1]) f[1][x] = 0;
}
int main() {
  int t = I();
  while (t-- > 0) {
    flandre();
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 I() {
  char c = getchar();
  long long x = 0;
  long long f = 1;
  while (c < '0' or c > '9') f = (c == '-') ? -1 : 1, c = getchar();
  while (c >= '0' and c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return ((f == 1) ? x : -x);
}
long long n, a[100005];
long long f[2][100005];
vector<long long> n0[2];
inline void ad(long long& x, long long y) { x = (x + y) % 998244353; }
void flandre() {
  n = I();
  for (long long i = 1; i <= n; ++i) a[i] = I();
  long long ans = 0;
  long long pre = 0, cur = 1;
  n0[0].clear();
  n0[1].clear();
  for (long long i = n; i >= 1; --i) {
    f[cur][a[i]] = 1;
    n0[cur].push_back(a[i]);
    long long lasp = -1;
    for (auto x : n0[pre]) {
      long long p = a[i] / ((a[i] + x - 1) / x);
      ad(f[cur][p], f[pre][x]);
      if (p != lasp and p != a[i]) n0[cur].push_back(p), lasp = p;
      ad(ans, f[pre][x] * i % 998244353 * ((a[i] - 1) / x) % 998244353);
    }
    for (auto x : n0[pre]) f[pre][x] = 0;
    n0[pre].clear();
    swap(pre, cur);
  }
  printf("%lld\n", ans);
  for (auto x : n0[0]) f[0][x] = 0;
  for (auto x : n0[1]) f[1][x] = 0;
}
int main() {
  int t = I();
  while (t-- > 0) {
    flandre();
  }
}

```