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>
#pragma GCC optimize(2)
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const int Mxdt = 100000, mod = 998244353;
char gc() {
  return getchar();
  static char buf[Mxdt], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, Mxdt, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
int read() {
  int a = 0;
  bool opt = 0;
  char c = gc();
  while (c < '0' || c > '9') opt |= c == '-', c = gc();
  while (c >= '0' && c <= '9') a = (a << 3) + (a << 1) + (c ^ 48), c = gc();
  return opt ? -a : a;
}
int Mod(int x) { return x < mod ? x : x - mod; }
void Add(int& x, int y) { x = Mod(x + y); }
int T, n, a[100005], s[2][100005], sz[2][100005], top[2], ans;
void ADD(int opt, int x, int num) {
  if (top[opt] && x == s[opt][top[opt]]) {
    Add(sz[opt][top[opt]], num);
  } else {
    s[opt][++top[opt]] = x;
    sz[opt][top[opt]] = num;
  }
}
int main() {
  T = read() + 1;
  while (--T) {
    n = read();
    ans = 0;
    for (int i = 1; i <= n; ++i) a[i] = read();
    int now = 0;
    top[now] = 0;
    for (int i = n; i; --i) {
      top[now ^ 1] = 0;
      for (int j = 1; j <= top[now]; ++j) {
        int x = (a[i] - 1) / (s[now][j]) + 1;
        Add(ans, 1ll * (x - 1) * sz[now][j] % mod * i % mod);
        ADD(now ^ 1, a[i] / x, sz[now][j]);
      }
      ADD(now ^ 1, a[i], 1);
      now ^= 1;
    }
    cout << ans << "\n";
  }
  return 0;
}
