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.
#include <bits/stdc++.h>
using namespace std;
namespace Ruri {
const int infi = 1e9;
const long long infl = 4e18;
}  // namespace Ruri
using namespace Ruri;
namespace Gadgets {
inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
}  // namespace Gadgets
namespace Read {
inline int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long rl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline double rd() {
  double x;
  scanf("%lf", &x);
  return x;
}
}  // namespace Read
namespace DeBug {}
using namespace Gadgets;
using namespace Read;
using namespace DeBug;
const int MAX_N = 1e5 + 5;
const int mod = 998244353;
void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void Minus(int &x, int y) {
  x -= y;
  x += (x >> 31) & mod;
};
void Mul(int &x, int y) { x = 1ll * x * y % mod; }
int n, a[MAX_N];
int dp[2][MAX_N];
vector<int> rec[2];
int main() {
  memset(dp, -1, sizeof(dp));
  int T = ri();
  while (T--) {
    rec[0].clear(), rec[1].clear();
    n = ri();
    for (int i = 1, bbb = n; i <= bbb; ++i) a[i] = ri();
    int ans = 0;
    int now = 1, pre = 0;
    dp[pre][a[n]] = 1;
    rec[pre].push_back(a[n]);
    for (int i = n - 1, bbb = 1; i >= bbb; --i) {
      for (auto val : rec[pre]) {
        int k = (a[i] + val - 1) / val,
            res = 1ll * i * (k - 1) % mod * dp[pre][val] % mod;
        Add(ans, res);
        int now_val = a[i] / k;
        if (dp[now][now_val] == -1)
          rec[now].push_back(now_val), dp[now][now_val] = 0;
        Add(dp[now][now_val], dp[pre][val]);
        dp[pre][val] = -1;
      }
      if (dp[now][a[i]] == -1) rec[now].push_back(a[i]), dp[now][a[i]] = 0;
      Add(dp[now][a[i]], 1), rec[pre].clear();
      swap(pre, now);
    }
    for (auto val : rec[pre]) dp[pre][val] = -1;
    printf("%lld\n", pos_mod(ans, mod));
  }
  return 0;
}
