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 target("avx")
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
using namespace std;
template <class T>
void rd(T &x) {
  char ch;
  x = 0;
  bool fl = false;
  while (!isdigit(ch = getchar())) (ch == '-') && (fl = true);
  for (x = (ch ^ '0'); isdigit(ch = getchar()); x = x * 10 + (ch ^ '0'))
    ;
  (fl == true) && (x = -x);
}
template <class T>
void output(T x) {
  if (x / 10) output(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
void ot(T x, int n) {
  for (auto i = 0; i < n; i++)
    i ? putchar('\n') : 0,
        (x[i] < 0 ? putchar('-'), output(-x[i]) : output(x[i]));
}
namespace Modulo {
using type = int;
const type mod = 998244353;
type ad(type x, type y) { return x + y >= mod ? x + y - mod : x + y; }
type sub(type x, type y) { return ad(x, mod - y); }
type mul(type x, type y) { return 1LL * x * y % mod; }
void inc(type &x, type y) { x = ad(x, y); }
void inc2(type &x, type y) { x = mul(x, y); }
type qm(type x, type y = mod - 2) {
  type ret = 1;
  while (y) {
    if (y & 1) ret = mul(x, ret);
    x = mul(x, x);
    y >>= 1;
  }
  return ret;
}
type div(type x, type y) { return mul(x, qm(y)); }
template <class... Args>
type ad(type a, type b, Args... args) {
  return ad(ad(a, b), args...);
}
template <class... Args>
type mul(type a, type b, Args... args) {
  return mul(mul(a, b), args...);
}
}  // namespace Modulo
using namespace Modulo;
map<int, int> mpp[100007];
int arr[100007];
int n;
int dfs(int i, int j) {
  if (i < 0) return 0;
  if (mpp[i].count(j)) return mpp[i][j];
  if (arr[i] % j == 0)
    return mpp[i][j] = ad(dfs(i - 1, j), mul(arr[i] / j - 1, i + 1));
  int tm = arr[i] / j + 1;
  return mpp[i][j] = ad(dfs(i - 1, arr[i] / tm), mul(tm - 1, i + 1));
}
void solve() {
  rd(n);
  for (auto _ = 0; _ < n; _++) rd(arr[_]);
  int ans = 0;
  for (auto i = n - 1; i > 0; i--)
    ans = ad(ans, dfs(i, arr[i])), mpp[i].clear();
  (ans < 0 ? putchar('-'), output(-ans) : output(ans));
  putchar('\n');
}
signed main() {
  int t = 1;
  rd(t);
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
