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("Ofast,unroll-loops,inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target( \
    "bmi,bmi2,lzcnt,popcnt,movbe,aes,pclmul,rdrnd,abm,mmx,avx,avx2,f16c,fma,sse,sse2,sse3,ssse3,sse4.1,sse4.2")
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 998244353;
const int MAXN = 200000;
void solve() {
  ll n, t = 0;
  cin >> n;
  int arr[n];
  for (int i = 0, nnnn = (n); i < nnnn; ++i) cin >> arr[i];
  vector<pair<int, int>> cur, nxt, nxt1;
  for (int i = n - 1; i >= 0; i--) {
    nxt.clear();
    nxt1.clear();
    for (auto it : cur) {
      t += (it.first - 1) * it.second * (i + 1);
      if (i > 0) {
        int t1 = arr[i] / it.first;
        int add = (arr[i - 1] + t1 - 1) / t1;
        if (nxt1.empty())
          nxt1.push_back(make_pair(add, it.second));
        else if (nxt1[((int)(nxt).size()) - 1].first == add) {
          nxt1[((int)(nxt).size()) - 1].second += it.second;
        } else {
          nxt1.push_back(make_pair(add, it.second));
        }
      }
    }
    t %= mod;
    if (i > 0) {
      int d = arr[i - 1] / arr[i] + (int)(arr[i - 1] % arr[i] > 0);
      pair<int, int> add = make_pair(d, 1);
      auto it = lower_bound(nxt1.begin(), nxt1.end(), add);
      for (auto it1 = nxt1.begin(); it1 != nxt1.end(); ++it1) {
        if (it == it1) {
          auto temp = *it1;
          if (temp.first == d) {
            temp.second++;
            nxt.push_back(temp);
          } else {
            nxt.push_back(add);
            nxt.push_back(temp);
          }
        } else
          nxt.push_back(*it1);
      }
      if (it == nxt1.end()) nxt.push_back(add);
    }
    cur = nxt;
  }
  cout << t << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}
