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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int64_t fx[10] = {0, 1, 0, -1, 1, -1, 1, -1};
int64_t fy[10] = {1, 0, -1, 0, 1, -1, -1, 1};
const long long M = 2e5 + 10, M2 = 2e6 + 10, oo = 1e9 + 7, mod = 998244353;
const long long MC = (1 << 20) + 5;
int64_t pos[M], dp[2][M];
void precomp() {}
void solve() {
  int64_t n;
  cin >> n;
  int64_t ara[n + 10];
  ara[0] = 0;
  for (int64_t i = 1; i <= n; i++) {
    cin >> ara[i];
  }
  ara[n + 1] = 0;
  long long ans = 0;
  for (int64_t i = n + 1; i >= 1; i--) {
    vector<int64_t> all;
    for (int64_t block = 1; block * block <= ara[i]; block++) {
      all.push_back(block);
      if (block < ara[i] / block) all.push_back(ara[i] / block);
    }
    if (i > 1) {
      dp[!(i & 1)][ara[i - 1]] = 1;
      for (int64_t b : all) {
        int64_t subs = dp[i & 1][b] * (i - 1),
                splits = ((ara[i - 1] + b - 1) / b) - 1;
        int64_t prev = ara[i - 1] / (splits + 1);
        dp[!(i & 1)][prev] += dp[i & 1][b];
        ans = (ans + (subs * splits) % mod) % mod;
      }
    }
    for (int64_t b : all) dp[i & 1][b] = 0;
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  precomp();
  int64_t T;
  cin >> T;
  for (int64_t tc = 1; tc <= T; tc++) {
    solve();
  }
  return 0;
}
