#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  register long long a = 0, po = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') po = -1, ch = getchar();
  while (isdigit(ch)) a = (a << 1) + (a << 3) + ch - 48, ch = getchar();
  return a * po;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
const long long maxn = 100010;
const long long mod = 998244353;
struct nowt {
  long long t, cy, cc;
};
void work() {
  vector<nowt> N, N1;
  long long n = read();
  long long tot = 0;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    a[i] = read();
  }
  N1.push_back({1, a[n - 1], n});
  for (long long i = n - 2; i >= 0; i--) {
    N.push_back({1, a[i], i + 1});
    long long lenn = (long long)N1.size();
    for (long long j = 0; j < (long long)N1.size(); j++) {
      long long t = (a[i] + N1[j].cy - 1) / N1[j].cy;
      long long d = a[i] / t;
      long long num = N.size();
      if (t != N[num - 1].t || d != N[num - 1].cy)
        N.push_back({t, d, N1[j].cc});
    }
    N.push_back({0, 0, n + 1});
    long long curr = (long long)N.size();
    for (long long j = 0; j < (long long)N.size() - 1; j++) {
      tot %= mod;
      tot += (i + 1) * (N[j].t - 1) % mod * (N[j + 1].cc - N[j].cc) % mod;
      tot %= mod;
    }
    N.pop_back();
    N1 = N;
    N.clear();
  }
  cout << tot << endl;
  a.clear();
}
signed main() {
  long long tt = read();
  while (tt--) {
    work();
  }
}

### Prompt

Please create a solution in cpp 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;
inline long long read() {
  register long long a = 0, po = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') po = -1, ch = getchar();
  while (isdigit(ch)) a = (a << 1) + (a << 3) + ch - 48, ch = getchar();
  return a * po;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
const long long maxn = 100010;
const long long mod = 998244353;
struct nowt {
  long long t, cy, cc;
};
void work() {
  vector<nowt> N, N1;
  long long n = read();
  long long tot = 0;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    a[i] = read();
  }
  N1.push_back({1, a[n - 1], n});
  for (long long i = n - 2; i >= 0; i--) {
    N.push_back({1, a[i], i + 1});
    long long lenn = (long long)N1.size();
    for (long long j = 0; j < (long long)N1.size(); j++) {
      long long t = (a[i] + N1[j].cy - 1) / N1[j].cy;
      long long d = a[i] / t;
      long long num = N.size();
      if (t != N[num - 1].t || d != N[num - 1].cy)
        N.push_back({t, d, N1[j].cc});
    }
    N.push_back({0, 0, n + 1});
    long long curr = (long long)N.size();
    for (long long j = 0; j < (long long)N.size() - 1; j++) {
      tot %= mod;
      tot += (i + 1) * (N[j].t - 1) % mod * (N[j + 1].cc - N[j].cc) % mod;
      tot %= mod;
    }
    N.pop_back();
    N1 = N;
    N.clear();
  }
  cout << tot << endl;
  a.clear();
}
signed main() {
  long long tt = read();
  while (tt--) {
    work();
  }
}

```