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;
const int MOD = 998244353;
const int inf = 0x3f3f3f3f;
const int N = 110000;
template <typename T>
inline bool scan_d(T& ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) {
    if ((c = getchar()) == EOF) return 0;
  }
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <typename T>
void print(T x) {
  static char s[33], *s1;
  s1 = s;
  if (!x) *s1++ = '0';
  if (x < 0) putchar('-'), x = -x;
  while (x) *s1++ = (x % 10 + '0'), x /= 10;
  while (s1-- != s) putchar(*s1);
}
template <typename T>
void println(T x) {
  print(x);
  putchar('\n');
}
int norm(int x) {
  if (x < 0) {
    x += MOD;
  }
  if (x >= MOD) {
    x -= MOD;
  }
  return x;
}
template <class T>
T power(T a, int b) {
  T res = 1;
  for (; b; b >>= 1, a *= a) {
    if (b & 1) {
      res *= a;
    }
  }
  return res;
}
struct mint {
  int x;
  mint(int x = 0) : x(norm(x)) {}
  int val() const { return x; }
  mint operator-() const { return mint(norm(MOD - x)); }
  mint inv() const { return power(*this, MOD - 2); }
  mint& operator*=(const mint& rv) {
    x = (long long)x * rv.x % MOD;
    return *this;
  }
  mint& operator+=(const mint& rv) {
    x = norm(x + rv.x);
    return *this;
  }
  mint& operator-=(const mint& rv) {
    x = norm(x - rv.x);
    return *this;
  }
  mint& operator/=(const mint& rv) { return *this *= rv.inv(); }
  friend mint operator*(const mint& lv, const mint& rv) {
    mint res = lv;
    res *= rv;
    return res;
  }
  friend mint operator+(const mint& lv, const mint& rv) {
    mint res = lv;
    res += rv;
    return res;
  }
  friend mint operator-(const mint& lv, const mint& rv) {
    mint res = lv;
    res -= rv;
    return res;
  }
  friend mint operator/(const mint& lv, const mint& rv) {
    mint res = lv;
    res /= rv;
    return res;
  }
};
struct TPoint {
  int v, x, y;
};
int n;
int a[N];
vector<TPoint> b[2];
vector<long long> dp[2];
void fmain() {
  scan_d(n);
  for (int i = (1); i < (n + 1); ++i) scan_d(a[i]);
  mint res = 0;
  int las = 0, cur = 0;
  for (int i = (1); i < (n + 1); ++i) {
    las = cur, cur ^= 1;
    b[cur].clear();
    for (int j = (1); j < (a[i] + 1); ++j) {
      if (j * j > a[i]) break;
      int v = j, x = a[i] / j, y = a[i] / j;
      if (a[i] % j != 0) y++;
      b[cur].push_back({v, x, y});
    }
    vector<TPoint> bb;
    for (int j = (0); j < (a[i] + 1); ++j) {
      if (j * j > a[i]) break;
      int v = (a[i] + j) / (j + 1), x = (j + 1), y = (j + 1);
      if (a[i] % (j + 1) != 0) x--;
      bb.push_back({v, x, y});
    }
    for (int j = (bb.size()) - 1; j >= (0); --j) b[cur].push_back(bb[j]);
    dp[cur].resize(b[cur].size());
    if (i == 1) {
      for (int j = (0); j < (dp[cur].size()); ++j) dp[cur][j] = b[cur][j].v - 1;
    } else {
      int k = 0;
      for (int j = (0); j < (dp[cur].size()); ++j) {
        while (b[las][k].y > b[cur][j].x) k++;
        dp[cur][j] = dp[las][k] + (long long)i * (b[cur][j].v - 1);
      }
    }
    res += (int)(dp[cur][0] % MOD);
  }
  println(res.val());
}
int main() {
  int T;
  scan_d(T);
  for (int t = (0); t < (T); ++t) {
    fmain();
  }
  return 0;
}
