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>
#pragma GCC optimize("O2")
using namespace std;
int dp[100002];
int l[100002];
int solven2(int n, vector<int> as);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tests;
  cin >> tests;
  for (int testid = 0; testid < tests; testid++) {
    vector<int> as;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      int k;
      cin >> k;
      as.push_back(k);
    }
    map<int, int> lvals;
    int sol = 0;
    for (int i = n - 1; i >= 0; i--) {
      map<int, int> newmap;
      for (auto rv : lvals) {
        int r = rv.first;
        int numSub = rv.second;
        int newOperationsFromr = INT_MAX;
        if (as[i] <= r) {
          newmap[as[i]] += numSub;
          newOperationsFromr = 0;
        } else {
          int lval = -1;
          newOperationsFromr = (as[i] - 1) / r;
          if (as[i] % r == 0)
            lval = r;
          else {
            lval = min(as[i] / (newOperationsFromr + 1), r - 1);
          }
          newmap[lval] += numSub;
        }
        long long tmp = (i + 1) * (newOperationsFromr % 998244353) * numSub;
        tmp %= 998244353;
        sol += int(tmp);
        sol %= 998244353;
      }
      newmap[as[i]]++;
      lvals = newmap;
    }
    cout << sol << "\n";
  }
}
int solven2(int n, vector<int> as) {
  if (n == 1) {
    return 0;
  }
  long long res = 0;
  for (int i = 1; i < n; i++) {
    if (as[i - 1] <= as[i]) {
      dp[i - 1] = 0;
      l[i - 1] = as[i - 1];
    } else {
      dp[i - 1] = (as[i - 1] - 1) / as[i];
      int lval = -1;
      if (as[i - 1] % as[i] == 0)
        lval = as[i];
      else {
        lval = min(as[i - 1] / (dp[i - 1] + 1), as[i] - 1);
      }
      l[i - 1] = lval;
    }
    res += dp[i - 1];
    res = res % 998244353;
  }
  for (int i = 2; i < n; i++) {
    for (int j = 0; j < n - i; j++) {
      if (as[j] <= l[j + 1]) {
        dp[j] = dp[j + 1];
        l[j] = as[j];
      } else {
        int newDivs = (as[j] - 1) / l[j + 1];
        dp[j] = dp[j + 1] + newDivs;
        int lval = -1;
        if (as[j] % l[j + 1] == 0)
          lval = l[j + 1];
        else {
          lval = min(as[j] / (newDivs + 1), l[j + 1] - 1);
        }
        l[j] = lval;
      }
      res += dp[j];
      res = res % 998244353;
    }
  }
  return res;
}
