#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5, m = 998244353;
long long bin_pow(long long a, long long b) {
  long long x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
}
long long readint() {
  long long x = 0, f = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar(), f |= c == '-';
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
string readstring() {
  string s = "";
  char c = getchar();
  while (!isalpha(c)) c = getchar();
  while (isalpha(c)) s += c, c = getchar();
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
void println(long long x) {
  print(x);
  putchar('\n');
}
void printsp(long long x) {
  print(x);
  putchar(' ');
}
void printyn(bool i) {
  if (i) {
    putchar('Y');
    putchar('E');
    putchar('S');
  } else {
    putchar('N');
    putchar('O');
  }
  putchar('\n');
};
struct dsu {
  long long p[maxn], r[maxn];
  long long n;
  dsu(long long n) {
    this->n = n;
    for (long long i = 0; i < n; ++i) p[i] = 1, r[i] = 0;
  }
  long long find(long long u) { return p[u] == u ? u : p[u] = find(p[u]); }
  bool unite(long long u, long long v) {
    u = find(u);
    v = find(v);
    if (r[u] < r[v]) swap(u, v);
    p[v] = u;
    r[u] += r[v];
  }
};
struct fw {
  long long t[maxn];
  long long n;
  fw(long long n) { this->n = n; }
  void upd(long long i, long long d) {
    for (; i < n; i |= i + 1) t[i] += d;
  }
  long long sum(long long i) {
    long long x = 0;
    for (; i >= 0; i &= i + 1, --i) x += t[i];
    return x;
  }
};
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a > b ? b : a; }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long len(long long x) {
  if (x > 0) return len(x / 10) + 1;
}
long long len2(long long x) {
  if (x == 1)
    return 1;
  else
    return len2(x - 1) * 10;
}
long long a[maxn], dp[2][maxn];
vector<long long> b[2];
signed main() {
  long long t = readint();
  while (t--) {
    long long n = readint(), ans = 0;
    for (long long i = 1; i <= n; ++i) a[i] = readint();
    for (long long i = n; i; --i) {
      long long j = i & 1, l = a[i];
      b[j].push_back(a[i]);
      dp[j][a[i]] = 1;
      for (long long x : b[j ^ 1]) {
        long long y = dp[j ^ 1][x], s = (a[i] + x - 1) / x, st = a[i] / s;
        ans += (s - 1) * y * i;
        dp[j][st] += y;
        if (l != st) b[j].push_back(st), l = st;
      }
      for (long long x : b[j ^ 1]) dp[j ^ 1][x] = 0;
      b[j ^ 1].clear();
      ans %= m;
    }
    println(ans);
    for (long long i = 0; i < 2; ++i)
      for (long long x : b[i]) dp[i][x] = 0;
    b[0].clear();
    b[1].clear();
  }
}

### Prompt

Create a solution in cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5, m = 998244353;
long long bin_pow(long long a, long long b) {
  long long x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
}
long long readint() {
  long long x = 0, f = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar(), f |= c == '-';
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
string readstring() {
  string s = "";
  char c = getchar();
  while (!isalpha(c)) c = getchar();
  while (isalpha(c)) s += c, c = getchar();
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
void println(long long x) {
  print(x);
  putchar('\n');
}
void printsp(long long x) {
  print(x);
  putchar(' ');
}
void printyn(bool i) {
  if (i) {
    putchar('Y');
    putchar('E');
    putchar('S');
  } else {
    putchar('N');
    putchar('O');
  }
  putchar('\n');
};
struct dsu {
  long long p[maxn], r[maxn];
  long long n;
  dsu(long long n) {
    this->n = n;
    for (long long i = 0; i < n; ++i) p[i] = 1, r[i] = 0;
  }
  long long find(long long u) { return p[u] == u ? u : p[u] = find(p[u]); }
  bool unite(long long u, long long v) {
    u = find(u);
    v = find(v);
    if (r[u] < r[v]) swap(u, v);
    p[v] = u;
    r[u] += r[v];
  }
};
struct fw {
  long long t[maxn];
  long long n;
  fw(long long n) { this->n = n; }
  void upd(long long i, long long d) {
    for (; i < n; i |= i + 1) t[i] += d;
  }
  long long sum(long long i) {
    long long x = 0;
    for (; i >= 0; i &= i + 1, --i) x += t[i];
    return x;
  }
};
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a > b ? b : a; }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long len(long long x) {
  if (x > 0) return len(x / 10) + 1;
}
long long len2(long long x) {
  if (x == 1)
    return 1;
  else
    return len2(x - 1) * 10;
}
long long a[maxn], dp[2][maxn];
vector<long long> b[2];
signed main() {
  long long t = readint();
  while (t--) {
    long long n = readint(), ans = 0;
    for (long long i = 1; i <= n; ++i) a[i] = readint();
    for (long long i = n; i; --i) {
      long long j = i & 1, l = a[i];
      b[j].push_back(a[i]);
      dp[j][a[i]] = 1;
      for (long long x : b[j ^ 1]) {
        long long y = dp[j ^ 1][x], s = (a[i] + x - 1) / x, st = a[i] / s;
        ans += (s - 1) * y * i;
        dp[j][st] += y;
        if (l != st) b[j].push_back(st), l = st;
      }
      for (long long x : b[j ^ 1]) dp[j ^ 1][x] = 0;
      b[j ^ 1].clear();
      ans %= m;
    }
    println(ans);
    for (long long i = 0; i < 2; ++i)
      for (long long x : b[i]) dp[i][x] = 0;
    b[0].clear();
    b[1].clear();
  }
}

```