#include <bits/stdc++.h>
long long q_pow(long long x, long long n, long long mod) {
  x %= mod;
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {1, 0};
int dy[] = {0, 1};
using namespace std;
void dbg_out() { cout << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
const long long maxn = 1e5 + 5;
const long long maxm = 250 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
vector<int> vec[2];
int dp[2][maxn];
int a[maxn];
int n;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int pre = 0, now = 1;
  long long ans = 0;
  long long seg, st;
  for (int i = n; i >= 1; i--) {
    swap(pre, now);
    dp[now][a[i]] = 1;
    vec[now].emplace_back(a[i]);
    int last = a[i];
    for (auto to : vec[pre]) {
      seg = (a[i] + to - 1) / to;
      st = a[i] / seg;
      ans += (seg - 1) * dp[pre][to] * i % mod;
      if (ans >= mod) ans -= mod;
      dp[now][st] += dp[pre][to];
      if (last != st) vec[now].emplace_back(st), last = st;
    }
    for (auto to : vec[pre]) {
      dp[pre][to] = 0;
    }
    vec[pre].clear();
  }
  for (auto to : vec[now]) {
    dp[now][to] = 0;
  }
  vec[now].clear();
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a cpp solution to 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>
long long q_pow(long long x, long long n, long long mod) {
  x %= mod;
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {1, 0};
int dy[] = {0, 1};
using namespace std;
void dbg_out() { cout << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
const long long maxn = 1e5 + 5;
const long long maxm = 250 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
vector<int> vec[2];
int dp[2][maxn];
int a[maxn];
int n;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int pre = 0, now = 1;
  long long ans = 0;
  long long seg, st;
  for (int i = n; i >= 1; i--) {
    swap(pre, now);
    dp[now][a[i]] = 1;
    vec[now].emplace_back(a[i]);
    int last = a[i];
    for (auto to : vec[pre]) {
      seg = (a[i] + to - 1) / to;
      st = a[i] / seg;
      ans += (seg - 1) * dp[pre][to] * i % mod;
      if (ans >= mod) ans -= mod;
      dp[now][st] += dp[pre][to];
      if (last != st) vec[now].emplace_back(st), last = st;
    }
    for (auto to : vec[pre]) {
      dp[pre][to] = 0;
    }
    vec[pre].clear();
  }
  for (auto to : vec[now]) {
    dp[now][to] = 0;
  }
  vec[now].clear();
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```