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 IO {
inline long long read() {
  long long w = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = (w << 1) + (w << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w * f;
}
}  // namespace IO
using namespace IO;
namespace CL {
const long long maxn = 1e5 + 5, mod = 998244353;
long long T, n, ans;
long long a[maxn], f[2][maxn];
bool vis[2][maxn];
vector<long long> vec[2];
inline void insert(long long cur, long long val) {
  if (vis[cur][val]) return;
  vis[cur][val] = 1, vec[cur].push_back(val);
}
inline void init() {
  for (long long i : vec[0]) f[0][i] = vis[0][i] = 0;
  for (long long i : vec[1]) f[1][i] = vis[1][i] = 0;
  vec[0].clear(), vec[1].clear();
  ans = 0;
}
inline long long main() {
  T = read();
  while (T--) {
    n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = n, cur = n & 1; i >= 1; i--, cur ^= 1) {
      f[cur][a[i]] = 1;
      insert(cur, a[i]);
      for (long long j : vec[cur ^ 1]) {
        long long k = (long long)ceil((double)a[i] / (double)j);
        f[cur][a[i] / k] = (f[cur][a[i] / k] + f[cur ^ 1][j]) % mod;
        insert(cur, a[i] / k);
        ans = (ans + i * f[cur ^ 1][j] % mod * (k - 1) % mod) % mod;
        f[cur ^ 1][j] = vis[cur ^ 1][j] = 0;
      }
      vec[cur ^ 1].clear();
    }
    printf("%lld\n", ans);
    init();
  }
  return 0;
}
}  // namespace CL
signed main() { return CL::main(); }
