#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-15;
const int MAXN = 100005;
const int LIM = 100000;
const double EPS = 1e-8;
const double PI = acos(-1.0);
const long long INF = 1e9 + 5;
const int MOD = 998244353;
int len = 62;
int a[MAXN];
vector<int> v[MAXN];
int t[MAXN];
int cnt = 0;
inline void block(int w) {
  for (int l = 1, r; l <= w; l = r + 1) {
    r = w / (w / l);
    v[w].push_back(l);
    cnt++;
  }
  v[w].push_back(w + 1);
}
inline void init() {
  for (int i = 1; i <= LIM; i++) block(i);
}
inline void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int ans = 0;
  vector<pair<int, int> > pre, cur;
  for (int i = n; i >= 1; i--) {
    int w = a[i] - 1;
    int edv = v[w].size() - 1;
    int edp = pre.size() - 1, tk = 0;
    for (int j = 1; j <= edv; j++) {
      int r = v[w][j] - 1, val = w / v[w][j - 1];
      int tmp = 0;
      while (tk <= edp && pre[tk].first <= r) {
        tmp += pre[tk].second;
        tmp %= MOD;
        tk++;
      }
      ans += (long long)tmp * val % MOD * i % MOD;
      ans %= MOD;
      if (tmp) cur.push_back({a[i] / (val + 1), tmp});
      if (tk > edp) break;
    }
    int tmp = 0;
    for (int j = 0; j <= edp; j++)
      if (pre[j].first >= a[i]) tmp += pre[j].second, tmp %= MOD;
    cur.push_back({a[i], (tmp + 1) % MOD});
    if (i == 1) break;
    pre = cur;
    cur.clear();
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-15;
const int MAXN = 100005;
const int LIM = 100000;
const double EPS = 1e-8;
const double PI = acos(-1.0);
const long long INF = 1e9 + 5;
const int MOD = 998244353;
int len = 62;
int a[MAXN];
vector<int> v[MAXN];
int t[MAXN];
int cnt = 0;
inline void block(int w) {
  for (int l = 1, r; l <= w; l = r + 1) {
    r = w / (w / l);
    v[w].push_back(l);
    cnt++;
  }
  v[w].push_back(w + 1);
}
inline void init() {
  for (int i = 1; i <= LIM; i++) block(i);
}
inline void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int ans = 0;
  vector<pair<int, int> > pre, cur;
  for (int i = n; i >= 1; i--) {
    int w = a[i] - 1;
    int edv = v[w].size() - 1;
    int edp = pre.size() - 1, tk = 0;
    for (int j = 1; j <= edv; j++) {
      int r = v[w][j] - 1, val = w / v[w][j - 1];
      int tmp = 0;
      while (tk <= edp && pre[tk].first <= r) {
        tmp += pre[tk].second;
        tmp %= MOD;
        tk++;
      }
      ans += (long long)tmp * val % MOD * i % MOD;
      ans %= MOD;
      if (tmp) cur.push_back({a[i] / (val + 1), tmp});
      if (tk > edp) break;
    }
    int tmp = 0;
    for (int j = 0; j <= edp; j++)
      if (pre[j].first >= a[i]) tmp += pre[j].second, tmp %= MOD;
    cur.push_back({a[i], (tmp + 1) % MOD});
    if (i == 1) break;
    pre = cur;
    cur.clear();
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```