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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long R = 998244353, R1 = 19491001, R2 = 236, NUMTESTCASE;
const long long NN = 10 + 1e5, Inverse2 = 500000004;
const double pi = acos(-1.0);
int di[8] = {1, 0, -1, 0, 1, -1, 1, -1}, dj[8] = {0, 1, 0, -1, 1, -1, -1, 1};
int n, Arr[NN];
pair<int, int> Dp[NN][2];
int Maps[2][NN], InvMaps[NN];
int main() {
  memset(InvMaps, -1, sizeof InvMaps);
  for (cin >> NUMTESTCASE; NUMTESTCASE; NUMTESTCASE--) {
    scanf("%d", &n);
    for (int i = (1); i <= (n); ++i) scanf("%d", Arr + i);
    int Ans = 0;
    for (long long i = 1; i <= n; i++)
      Ans = (Ans + 1ll * (n - i + 2) * (n - i + 1) / 2 % R) % R;
    Ans = (R - Ans) % R;
    for (int i = (n); i >= (1); --i) {
      InvMaps[Arr[i]] = 1;
      Maps[i & 1][0] = 1;
      Maps[i & 1][1] = Arr[i];
      Dp[Arr[i]][i & 1] = {1, 1};
      Ans = (Ans + 1) % R;
      for (int k = 1; k <= Maps[i & 1 ^ 1][0]; k++) {
        pair<int, pair<int, int>> p = {Maps[i & 1 ^ 1][k],
                                       Dp[Maps[i & 1 ^ 1][k]][i & 1 ^ 1]};
        int t = 0, Nxt = Arr[i], Cnt = p.second.second;
        if (Arr[i] <= p.first)
          t = 1;
        else
          t = (Arr[i] + p.first - 1) / p.first,
          Nxt = Arr[i] / ((Arr[i] + p.first - 1) / p.first);
        t = (1ll * t * Cnt % R + p.second.first) % R;
        Ans = (Ans + t) % R;
        int &Pos = InvMaps[Nxt];
        if (Pos == -1)
          Pos = ++Maps[i & 1][0], Maps[i & 1][Maps[i & 1][0]] = Nxt;
        Dp[Nxt][i & 1] = {(Dp[Nxt][i & 1].first + t) % R,
                          Dp[Nxt][i & 1].second + Cnt};
      }
      for (int k = (1); k <= (Maps[i & 1 ^ 1][0]); ++k)
        InvMaps[Maps[i & 1 ^ 1][k]] = -1,
                             Dp[Maps[i & 1 ^ 1][k]][i & 1 ^ 1] = {0, 0};
    }
    for (int k = (1); k <= (Maps[n & 1][0]); ++k)
      InvMaps[Maps[n & 1][k]] = -1, Dp[Maps[n & 1][k]][n & 1] = {0, 0};
    printf("%d\n", Ans);
  }
  return 0;
}
