#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << "\n"; }
template <typename H, typename... T>
void debug_out(H h, T... t) {
  cerr << " " << (h);
  debug_out(t...);
}
const int MOD = 998244353;
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long add(long long a, long long b) { return (a + b) % MOD; }
const int N = 1e5 + 2;
const int S = 702;
int b[2][S];
int siz[N];
int dp[2][S];
int a[N];
void init(int i) {
  auto &bb = b[i % 2];
  int x = a[i];
  siz[i] = 0;
  for (int k = 1; k <= min(S, x); ++k) {
    if (siz[i] and x / k == bb[siz[i] - 1]) break;
    bb[siz[i]++] = x / k;
  }
  for (int val = 1; val <= min(S, x); ++val) {
    if (x / (x / val) != val) break;
    bb[siz[i]++] = val;
  }
  sort(bb, bb + siz[i]);
  siz[i] = unique(bb, bb + siz[i]) - bb;
}
void solve() {
  int n, res;
  cin >> n;
  fill(siz, siz + n + 1, 0);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  fill(dp[0], dp[0] + S, 0);
  fill(dp[1], dp[1] + S, 0);
  res = 0;
  init(n);
  dp[n % 2][siz[n]] = 1;
  int sz, x, y, ad, k, pos, s, idx;
  for (int i = n - 1; i >= 1; --i) {
    init(i);
    auto &b1 = b[(i + 1) % 2];
    sz = siz[i + 1];
    auto &bb = b[i % 2];
    s = siz[i];
    auto &cur = dp[i % 2];
    fill(cur, cur + S, 0);
    cur[s] = 1;
    auto &nxt = dp[(i + 1) % 2];
    idx = 0;
    for (int j = 0; j < sz; ++j) {
      ad = nxt[j + 1];
      if (ad == 0) continue;
      y = b1[j];
      if (a[i] <= y) {
        cur[s] = add(cur[s], ad);
        continue;
      }
      k = (a[i] + y - 1) / y;
      x = a[i] / k;
      while (idx < s and bb[idx] < x) ++idx;
      assert(idx < s and bb[idx] == x);
      pos = idx + 1;
      cur[pos] = add(cur[pos], ad);
      res = add(res, mul(k - 1, mul(ad, i)));
    }
  }
  cout << res << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void debug_out() { cerr << "\n"; }
template <typename H, typename... T>
void debug_out(H h, T... t) {
  cerr << " " << (h);
  debug_out(t...);
}
const int MOD = 998244353;
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long add(long long a, long long b) { return (a + b) % MOD; }
const int N = 1e5 + 2;
const int S = 702;
int b[2][S];
int siz[N];
int dp[2][S];
int a[N];
void init(int i) {
  auto &bb = b[i % 2];
  int x = a[i];
  siz[i] = 0;
  for (int k = 1; k <= min(S, x); ++k) {
    if (siz[i] and x / k == bb[siz[i] - 1]) break;
    bb[siz[i]++] = x / k;
  }
  for (int val = 1; val <= min(S, x); ++val) {
    if (x / (x / val) != val) break;
    bb[siz[i]++] = val;
  }
  sort(bb, bb + siz[i]);
  siz[i] = unique(bb, bb + siz[i]) - bb;
}
void solve() {
  int n, res;
  cin >> n;
  fill(siz, siz + n + 1, 0);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  fill(dp[0], dp[0] + S, 0);
  fill(dp[1], dp[1] + S, 0);
  res = 0;
  init(n);
  dp[n % 2][siz[n]] = 1;
  int sz, x, y, ad, k, pos, s, idx;
  for (int i = n - 1; i >= 1; --i) {
    init(i);
    auto &b1 = b[(i + 1) % 2];
    sz = siz[i + 1];
    auto &bb = b[i % 2];
    s = siz[i];
    auto &cur = dp[i % 2];
    fill(cur, cur + S, 0);
    cur[s] = 1;
    auto &nxt = dp[(i + 1) % 2];
    idx = 0;
    for (int j = 0; j < sz; ++j) {
      ad = nxt[j + 1];
      if (ad == 0) continue;
      y = b1[j];
      if (a[i] <= y) {
        cur[s] = add(cur[s], ad);
        continue;
      }
      k = (a[i] + y - 1) / y;
      x = a[i] / k;
      while (idx < s and bb[idx] < x) ++idx;
      assert(idx < s and bb[idx] == x);
      pos = idx + 1;
      cur[pos] = add(cur[pos], ad);
      res = add(res, mul(k - 1, mul(ad, i)));
    }
  }
  cout << res << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

```