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;
vector<long long> a;
string s;
long long n;
long long dp[100000];
long long memo[100000][50 + 1];
long long find(long long pos, long long val) {
  if (pos < 0) return 0;
  if (a[pos - 1] <= val) return dp[pos - 1];
  if (val <= 50 && memo[pos][val] != -1) return memo[pos][val];
  long long x = a[pos - 1] / val;
  if (a[pos - 1] % val) x++;
  long long nex = a[pos - 1] / x;
  long long cnt = x - 1;
  long long ans = (find(pos - 1, nex) + (cnt * (pos)) % 998244353) % 998244353;
  if (val <= 50) memo[pos][val] = ans;
  return ans;
}
long long solve() {
  a.clear();
  s.clear();
  long long i = 0, j = 0, k = 0, x = 0, y = 0, z = 0;
  cin >> n;
  a.resize(n);
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    dp[i] = 0;
    for (j = 0; j <= 50; j++) memo[i][j] = -1;
  }
  long long ans = 0;
  for (i = 1; i < n; i++) {
    dp[i] = find(i, a[i]);
    ans = (ans + dp[i]) % 998244353;
  }
  cout << ans << "\n";
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long testcases = 1;
  cin >> testcases;
  while (testcases--) {
    solve();
  }
  return 0;
}
