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;
template <typename T>
inline void ckmax(T &x, T y) {
  x = x < y ? y : x;
  return;
}
template <typename T>
inline void ckmin(T &x, T y) {
  x = x < y ? x : y;
  return;
}
namespace FastIO {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
char OPUT[100];
inline void print(long long x) {
  if (!x) return putchar('0'), putchar('\n'), void();
  if (x < 0) putchar('-'), x = -x;
  long long cnt = 0;
  while (x) OPUT[++cnt] = x % 10, x /= 10;
  for (long long i = cnt; i >= 1; --i) putchar(OPUT[i] + '0');
  putchar('\n');
  return;
}
}  // namespace FastIO
using namespace FastIO;
const long long mod = 998244353;
inline long long add(long long x, long long y, long long Mod = mod) {
  return x + y >= Mod ? x + y - Mod : x + y;
}
inline long long del(long long x, long long y, long long Mod = mod) {
  return x - y < 0 ? x - y + Mod : x - y;
}
inline long long mul(long long x, long long y, long long Mod = mod) {
  return x * y - x * y / Mod * Mod;
}
inline void ckadd(long long &x, long long y, long long Mod = mod) {
  x = x + y >= Mod ? x + y - Mod : x + y;
}
inline void ckdel(long long &x, long long y, long long Mod = mod) {
  x = x - y < 0 ? x - y + Mod : x - y;
}
inline void ckmul(long long &x, long long y, long long Mod = mod) {
  x = x * y - x * y / Mod * Mod;
}
inline long long ksm(long long x, long long y, long long Mod = mod) {
  long long res = 1;
  for (; y; y >>= 1, ckmul(x, x, Mod))
    if (y & 1) ckmul(res, x, Mod);
  return res;
}
inline void approx(long long val, long long Mod = mod, long long lim = 1e5) {
  long long x = val, y = Mod, a = 1, b = 0;
  while (x > lim) {
    swap(x, y);
    swap(a, b);
    a -= x / y * b;
    x %= y;
  }
  cout << x << "/" << a << endl;
  return;
}
const long long N = 1e5 + 10;
long long dp[2][N], vis[2][N], a[N], n;
vector<long long> app[2];
inline void insert(long long cur, long long val) {
  if (!vis[cur][val]) vis[cur][val] = 1, app[cur].push_back(val);
  return;
}
signed main() {
  long long T = read();
  while (T--) {
    n = read();
    for (long long i = 1; i <= n; ++i) a[i] = read();
    long long cur = 0, ans = 0;
    for (long long i = n; i >= 1; --i, cur ^= 1) {
      ckadd(dp[cur][a[i]], 1);
      insert(cur, a[i]);
      for (auto t : app[cur ^ 1]) {
        long long k = (a[i] + t - 1) / t;
        ckadd(dp[cur][a[i] / k], dp[cur ^ 1][t]);
        insert(cur, a[i] / k);
        ckadd(ans, mul(i, mul(dp[cur ^ 1][t], k - 1)));
        dp[cur ^ 1][t] = vis[cur ^ 1][t] = 0;
      }
      app[cur ^ 1].clear();
    }
    cur ^= 1;
    for (auto t : app[cur]) vis[cur][t] = dp[cur][t] = 0;
    app[cur].clear();
    print(ans);
  }
  return 0;
}
