#include <bits/stdc++.h>
using namespace std;
void print_err() { cerr << "\n"; }
template <class T, class... Arg>
void print_err(T x, Arg &&...args) {
  cerr << x << " ";
  print_err(args...);
}
template <class T>
void print_container(T &cont) {
  for (auto iter : cont) {
    cerr << iter << " ";
  }
  cerr << "\n";
}
int mod = 998244353;
long long ceil(long long a, long long b) { return (a + b - 1) / b; }
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
vector<int> d4x{0, 0, 1, -1};
vector<int> d4y{-1, 1, 0, 0};
vector<int> d8x{0, 0, 1, -1, 1, 1, -1, -1};
vector<int> d8y{1, -1, 0, 0, -1, 1, 1, -1};
vector<long long> B(1e5 + 1, -1), C(1e5 + 1, -1);
void solve(int test) {
  ;
  int n;
  cin >> n;
  vector<int> A(n);
  for (int &a : A) {
    cin >> a;
  }
  long long ans = 0;
  vector<long long> dp1, dp2;
  int j;
  long long val;
  for (int i = n - 1; i >= 0; --i) {
    for (auto iter : dp1) {
      j = iter;
      val = B[j];
      if (val < 0) {
        ;
      }
      if (val > 0) {
        if (j >= A[i]) {
          if (C[A[i]] == -1) {
            dp2.push_back(A[i]);
            C[A[i]] = val;
            C[A[i]] %= mod;
          } else {
            C[A[i]] += val;
            C[A[i]] %= mod;
          }
        } else {
          if (C[A[i] / ceil(A[i], j)] == -1) {
            dp2.push_back(A[i] / ceil(A[i], j));
            C[A[i] / ceil(A[i], j)] = val;
            ans += (i + 1) * (ceil(A[i], j) - 1) * val;
            ans %= mod;
            C[A[i] / ceil(A[i], j)] %= mod;
          } else {
            C[A[i] / ceil(A[i], j)] += val;
            ans += (i + 1) * (ceil(A[i], j) - 1) * val;
            ans %= mod;
            C[A[i] / ceil(A[i], j)] %= mod;
          }
        }
      }
      B[j] = -1;
    }
    dp1.clear();
    swap(dp1, dp2);
    swap(B, C);
    if (B[A[i]] == -1) {
      dp1.push_back(A[i]);
      B[A[i]] = 1;
    } else {
      B[A[i]]++;
      B[A[i]] %= mod;
    }
  }
  for (int j : dp1) {
    B[j] = -1;
  }
  cout << ans << endl;
}
int main() {
  int i = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test_cases;
  cin >> test_cases;
  for (i = 1; i <= test_cases; ++i) solve(i);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
void print_err() { cerr << "\n"; }
template <class T, class... Arg>
void print_err(T x, Arg &&...args) {
  cerr << x << " ";
  print_err(args...);
}
template <class T>
void print_container(T &cont) {
  for (auto iter : cont) {
    cerr << iter << " ";
  }
  cerr << "\n";
}
int mod = 998244353;
long long ceil(long long a, long long b) { return (a + b - 1) / b; }
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
vector<int> d4x{0, 0, 1, -1};
vector<int> d4y{-1, 1, 0, 0};
vector<int> d8x{0, 0, 1, -1, 1, 1, -1, -1};
vector<int> d8y{1, -1, 0, 0, -1, 1, 1, -1};
vector<long long> B(1e5 + 1, -1), C(1e5 + 1, -1);
void solve(int test) {
  ;
  int n;
  cin >> n;
  vector<int> A(n);
  for (int &a : A) {
    cin >> a;
  }
  long long ans = 0;
  vector<long long> dp1, dp2;
  int j;
  long long val;
  for (int i = n - 1; i >= 0; --i) {
    for (auto iter : dp1) {
      j = iter;
      val = B[j];
      if (val < 0) {
        ;
      }
      if (val > 0) {
        if (j >= A[i]) {
          if (C[A[i]] == -1) {
            dp2.push_back(A[i]);
            C[A[i]] = val;
            C[A[i]] %= mod;
          } else {
            C[A[i]] += val;
            C[A[i]] %= mod;
          }
        } else {
          if (C[A[i] / ceil(A[i], j)] == -1) {
            dp2.push_back(A[i] / ceil(A[i], j));
            C[A[i] / ceil(A[i], j)] = val;
            ans += (i + 1) * (ceil(A[i], j) - 1) * val;
            ans %= mod;
            C[A[i] / ceil(A[i], j)] %= mod;
          } else {
            C[A[i] / ceil(A[i], j)] += val;
            ans += (i + 1) * (ceil(A[i], j) - 1) * val;
            ans %= mod;
            C[A[i] / ceil(A[i], j)] %= mod;
          }
        }
      }
      B[j] = -1;
    }
    dp1.clear();
    swap(dp1, dp2);
    swap(B, C);
    if (B[A[i]] == -1) {
      dp1.push_back(A[i]);
      B[A[i]] = 1;
    } else {
      B[A[i]]++;
      B[A[i]] %= mod;
    }
  }
  for (int j : dp1) {
    B[j] = -1;
  }
  cout << ans << endl;
}
int main() {
  int i = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test_cases;
  cin >> test_cases;
  for (i = 1; i <= test_cases; ++i) solve(i);
  return 0;
}

```