#include <bits/stdc++.h>
using namespace std;
using namespace std;
inline int read() {
  char ch;
  bool sign = 0;
  int res = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = 1;
  while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
  if (sign)
    return -res;
  else
    return res;
}
const int maxn = 100000 + 10;
const int mod = 998244353;
const int inf = 1e9;
const int maxm = 640;
int a[maxn], n;
vector<int> val;
inline int add(int x, int y) {
  x += y;
  if (x >= mod)
    x -= mod;
  else if (x < 0)
    x += mod;
  return x;
}
inline int mul(int x, int y) { return 1LL * x * y % mod; }
int dp[2][maxm];
inline void clear(int t) {
  for (int i = (0); i < (maxm); ++i) {
    dp[t][i] = 0;
  }
}
inline int solve() {
  int t = 0;
  clear(t);
  int ans = 0;
  val.clear();
  for (int l = 1; l <= a[1];) {
    val.push_back(a[1] / l);
    l = a[1] / val.back() + 1;
  }
  for (int i = (2); i <= (n); ++i) {
    clear(t ^ 1);
    int now = 0;
    vector<int> nowval;
    int l = 1;
    int c = 0;
    for (; l <= a[i];) {
      nowval.push_back(a[i] / l);
      int j = nowval[c];
      l = a[i] / j + 1;
      int tmp = a[i - 1] / ((a[i - 1] + j - 1) / j);
      for (; val[now] != tmp && now >= 0;) {
        ++now;
      }
      assert(now >= 0);
      assert(val[now] == tmp);
      int v = a[i - 1] / j;
      if (a[i - 1] % j == 0) v--;
      dp[t ^ 1][c] = add(dp[t][now], mul(i - 1, v));
      c++;
    }
    ans = add(ans, dp[t ^ 1][0]);
    val = nowval;
    t ^= 1;
  }
  return ans;
}
int main(void) {
  int T;
  scanf("%d", &(T));
  for (; T--;) {
    scanf("%d", &(n));
    for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
    printf("%d\n", solve());
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using namespace std;
inline int read() {
  char ch;
  bool sign = 0;
  int res = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = 1;
  while (isdigit(ch)) res = res * 10 + (ch ^ 48), ch = getchar();
  if (sign)
    return -res;
  else
    return res;
}
const int maxn = 100000 + 10;
const int mod = 998244353;
const int inf = 1e9;
const int maxm = 640;
int a[maxn], n;
vector<int> val;
inline int add(int x, int y) {
  x += y;
  if (x >= mod)
    x -= mod;
  else if (x < 0)
    x += mod;
  return x;
}
inline int mul(int x, int y) { return 1LL * x * y % mod; }
int dp[2][maxm];
inline void clear(int t) {
  for (int i = (0); i < (maxm); ++i) {
    dp[t][i] = 0;
  }
}
inline int solve() {
  int t = 0;
  clear(t);
  int ans = 0;
  val.clear();
  for (int l = 1; l <= a[1];) {
    val.push_back(a[1] / l);
    l = a[1] / val.back() + 1;
  }
  for (int i = (2); i <= (n); ++i) {
    clear(t ^ 1);
    int now = 0;
    vector<int> nowval;
    int l = 1;
    int c = 0;
    for (; l <= a[i];) {
      nowval.push_back(a[i] / l);
      int j = nowval[c];
      l = a[i] / j + 1;
      int tmp = a[i - 1] / ((a[i - 1] + j - 1) / j);
      for (; val[now] != tmp && now >= 0;) {
        ++now;
      }
      assert(now >= 0);
      assert(val[now] == tmp);
      int v = a[i - 1] / j;
      if (a[i - 1] % j == 0) v--;
      dp[t ^ 1][c] = add(dp[t][now], mul(i - 1, v));
      c++;
    }
    ans = add(ans, dp[t ^ 1][0]);
    val = nowval;
    t ^= 1;
  }
  return ans;
}
int main(void) {
  int T;
  scanf("%d", &(T));
  for (; T--;) {
    scanf("%d", &(n));
    for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
    printf("%d\n", solve());
  }
  return 0;
}

```