#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
long long cas, n, a[100005], t[2][100005], b[2][100005], tot[2], ans,
    sum[2][100005];
long long ceil(long long x, long long y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
signed main() {
  cas = read();
  while (cas--) {
    n = read();
    for (long long i = 1; i <= n; ++i) a[i] = read();
    ans = 0;
    for (long long i = n, r = 1; i >= 1; --i, r ^= 1) {
      for (long long j = 1; j <= tot[r ^ 1]; ++j) {
        (ans +=
         t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353 +
         sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        (sum[r ^ 1][b[r ^ 1][j]] +=
         t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353) %=
            998244353;
        long long tr = a[i] / ceil(a[i], b[r ^ 1][j]);
        if (!t[r][tr]) b[r][++tot[r]] = tr;
        (t[r][tr] += t[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        (sum[r][tr] += sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        t[r ^ 1][b[r ^ 1][j]] = 0;
        sum[r ^ 1][b[r ^ 1][j]] = 0;
      }
      tot[r ^ 1] = 0;
      if (!t[r][a[i]]) b[r][++tot[r]] = a[i];
      t[r][a[i]]++;
    }
    printf("%d\n", (ans % 998244353 + 998244353) % 998244353);
    for (long long i = 1; i <= tot[n & 1]; ++i) {
      t[n & 1][b[n & 1][i]] = 0;
      sum[n & 1][b[n & 1][i]] = 0;
    }
    tot[n & 1] = 0;
  }
  return 0;
}

### Prompt

Generate 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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
long long cas, n, a[100005], t[2][100005], b[2][100005], tot[2], ans,
    sum[2][100005];
long long ceil(long long x, long long y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
signed main() {
  cas = read();
  while (cas--) {
    n = read();
    for (long long i = 1; i <= n; ++i) a[i] = read();
    ans = 0;
    for (long long i = n, r = 1; i >= 1; --i, r ^= 1) {
      for (long long j = 1; j <= tot[r ^ 1]; ++j) {
        (ans +=
         t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353 +
         sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        (sum[r ^ 1][b[r ^ 1][j]] +=
         t[r ^ 1][b[r ^ 1][j]] * (ceil(a[i], b[r ^ 1][j]) - 1) % 998244353) %=
            998244353;
        long long tr = a[i] / ceil(a[i], b[r ^ 1][j]);
        if (!t[r][tr]) b[r][++tot[r]] = tr;
        (t[r][tr] += t[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        (sum[r][tr] += sum[r ^ 1][b[r ^ 1][j]]) %= 998244353;
        t[r ^ 1][b[r ^ 1][j]] = 0;
        sum[r ^ 1][b[r ^ 1][j]] = 0;
      }
      tot[r ^ 1] = 0;
      if (!t[r][a[i]]) b[r][++tot[r]] = a[i];
      t[r][a[i]]++;
    }
    printf("%d\n", (ans % 998244353 + 998244353) % 998244353);
    for (long long i = 1; i <= tot[n & 1]; ++i) {
      t[n & 1][b[n & 1][i]] = 0;
      sum[n & 1][b[n & 1][i]] = 0;
    }
    tot[n & 1] = 0;
  }
  return 0;
}

```