#include <bits/stdc++.h>
#pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
const int MAXN = 1e5 + 100;
int A[MAXN];
const int MOD = 998244353;
int mp1[MAXN];
int mp2[MAXN];
bool mk1[MAXN];
bool mk2[MAXN];
vector<int> v1, v2;
void limpiar1() {
  for (auto x : v1) {
    mk1[x] = false;
    mp1[x] = 0;
  }
  v1.clear();
}
void limpiar2() {
  for (auto x : v2) {
    mk2[x] = false;
    mp2[x] = 0;
  }
  v2.clear();
}
void sumar1(int x, int v) {
  if (!mk1[x]) mk1[x] = true, v1.push_back(x);
  mp1[x] += v;
}
void sumar2(int x, int v) {
  if (!mk2[x]) mk2[x] = true, v2.push_back(x);
  mp2[x] += v;
}
void copiar() {
  for (auto x : v2) {
    sumar1(x, mp2[x]);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int N;
    cin >> N;
    for (int i = 1; i <= N; i++) {
      cin >> A[i];
    }
    int res = 0;
    for (int i = N; i >= 1; i--) {
      for (auto x : v1) {
        if (A[i] <= x) {
          sumar2(A[i], mp1[x]);
        } else if (A[i] % x == 0) {
          sumar2(x, mp1[x]);
          res += (long long)(A[i] / x - 1) * (long long)mp1[x] % MOD *
                 (long long)i % MOD;
          res %= MOD;
        } else {
          sumar2(A[i] / (A[i] / x + 1), mp1[x]);
          res += (long long)(A[i] / x) * (long long)mp1[x] % MOD *
                 (long long)i % MOD;
          res %= MOD;
        }
      }
      sumar2(A[i], 1);
      limpiar1();
      copiar();
      limpiar2();
    }
    limpiar1();
    cout << res << '\n';
  }
  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>
#pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
const int MAXN = 1e5 + 100;
int A[MAXN];
const int MOD = 998244353;
int mp1[MAXN];
int mp2[MAXN];
bool mk1[MAXN];
bool mk2[MAXN];
vector<int> v1, v2;
void limpiar1() {
  for (auto x : v1) {
    mk1[x] = false;
    mp1[x] = 0;
  }
  v1.clear();
}
void limpiar2() {
  for (auto x : v2) {
    mk2[x] = false;
    mp2[x] = 0;
  }
  v2.clear();
}
void sumar1(int x, int v) {
  if (!mk1[x]) mk1[x] = true, v1.push_back(x);
  mp1[x] += v;
}
void sumar2(int x, int v) {
  if (!mk2[x]) mk2[x] = true, v2.push_back(x);
  mp2[x] += v;
}
void copiar() {
  for (auto x : v2) {
    sumar1(x, mp2[x]);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int N;
    cin >> N;
    for (int i = 1; i <= N; i++) {
      cin >> A[i];
    }
    int res = 0;
    for (int i = N; i >= 1; i--) {
      for (auto x : v1) {
        if (A[i] <= x) {
          sumar2(A[i], mp1[x]);
        } else if (A[i] % x == 0) {
          sumar2(x, mp1[x]);
          res += (long long)(A[i] / x - 1) * (long long)mp1[x] % MOD *
                 (long long)i % MOD;
          res %= MOD;
        } else {
          sumar2(A[i] / (A[i] / x + 1), mp1[x]);
          res += (long long)(A[i] / x) * (long long)mp1[x] % MOD *
                 (long long)i % MOD;
          res %= MOD;
        }
      }
      sumar2(A[i], 1);
      limpiar1();
      copiar();
      limpiar2();
    }
    limpiar1();
    cout << res << '\n';
  }
  return 0;
}

```