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.
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  int c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long t, n, ans, lst, a[100009], b1[100009], b2[100009], c[100009],
    d[100009];
void calc(long long x, long long y) {
  for (int i = y + 1; i <= x; i++)
    d[i] = (d[i - 1] + c[i] * i % 998244353) % 998244353,
    ans += c[i] * i % 998244353, ans %= 998244353;
}
void solve() {
  n = read();
  for (int i = 1; i <= n; i++) b1[i] = b2[i] = c[i] = d[i] = 0;
  for (int i = 1; i <= n; i++) a[i] = read();
  ans = 0;
  for (int i = 1; i <= n; i++) {
    lst = a[i];
    int j;
    b1[i] = b2[i] = a[i];
    for (j = i - 1; j >= 1; j--) {
      if (b2[j] > lst) {
        c[j] = (a[j] + lst - 1) / lst;
        b1[j] = a[j] / c[j];
        if (a[j] % c[j] == 0)
          b2[j] = b1[j];
        else
          b2[j] = b1[j] + 1;
        lst = b1[j], c[j]--;
      } else {
        ans += d[j], ans %= 998244353;
        break;
      }
    }
    calc(i, j);
  }
}
int main() {
  t = read();
  while (t--) solve(), printf("%lld\n", ans);
  return 0;
}
