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 optimize(3)
using namespace std;
const int maxn = 1e5 + 11, maxm = 1e5 + 11, inf_int = 0x3f3f3f3f;
const long long inf_ll = 0x3f3f3f3f3f3f3f, mod = 998244353;
const double eps = 1e-6;
const long long inv2 = (mod + 1) / 2;
struct custom_hash {
  static size_t splitmix64(size_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(size_t x) const {
    static const size_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
unordered_map<int, int, custom_hash> f[2];
int n, a[maxn], pre, cur;
long long ans, en[maxn][100];
inline long long DP(long long x, long long v) {
  if (x == 0) return 0;
  if (x > 1 && f[pre].count(a[x] / ((a[x - 1] + v - 1) / v)))
    return f[pre][a[x] / ((a[x - 1] + v - 1) / v)];
  long long res = 0, tv = v;
  for (int i = x; i; i--)
    if (v < 100) {
      res = (res + en[i][v]) % mod;
      break;
    } else {
      int t0 = (a[i] + v - 1) / v;
      res = (res + 1ll * i * (t0 - 1) % mod) % mod;
      v = a[i] / t0;
    }
  return f[cur][tv] = res;
}
inline long long dfs(int x, int v) {
  if (x == 0) return 0;
  if (en[x][v] != -1) return en[x][v];
  int t0 = (a[x] + v - 1) / v;
  en[x][v] = (dfs(x - 1, a[x] / t0) + 1ll * x * (t0 - 1) % mod) % mod;
  return en[x][v];
}
inline void solve() {
  cin >> n, ans = 0, cur = 1, pre = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 1; j < 100; j++) en[i][j] = -1;
  }
  for (int i = 1; i < n; i++)
    for (int j = 1; j < 100; j++) dfs(i, j);
  f[pre].clear();
  for (int i = 1; i < n; i++) {
    f[cur].clear();
    ans = (ans + DP(i, a[i + 1])) % mod;
    swap(cur, pre);
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int TEST = 1;
  cin >> TEST;
  while (TEST--) solve();
}
