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>
#include<ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define mp make_pair
#define mt make_tuple
#define all(x) x.begin(), x.end()
#define ff first
#define ss second
using namespace std;
template <typename T>
using matrix = vector<vector<T>>;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll INFL = (1LL<<62)-1;
const int INF = (1<<30)-1;
const double EPS = 1e-7;
const int MOD = 998244353;
const int MAXN = 1e6+1;
const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();

struct chash{
    int operator() (int id) const{
        return id^RANDOM;
    }
};
int main(){

    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        vector<int> v(n);
        for(int i = 0; i < n; i++)
            cin >> v[i];
        

        ll resp = 0;
        vector<ll> dp ={0};
        vector<int> val={1};
        for(int i = 0; i < n-1; i++){
            
            vector<int> div, div2;
            div.reserve(sqrt(v[i])*2),div2.reserve(sqrt(v[i]));
            for(int j = 1; j <= sqrt(v[i+1]); j++){
                div.push_back(j), div2.push_back(v[i+1]/j);
            }
            for(int j = ssize(div2)-1; j >= 0; j--)
                div.push_back(div2[j]);
            div.erase(unique(all(div)),div.end());

            vector<ll> newdp(div.size());
            int ptr = 0;
            int check = ssize(dp)-1;
            for(int j = 0; j < div.size(); j++){
                ll group = (v[i]+div[j]-1)/(div[j]);
                int x = v[i]/group;
                while(ptr < ssize(val)-1 && val[ptr+1] <= x){
                    ptr++;
                }
                
                newdp[j] = ((group-1)*(i+1)+dp[ptr])%MOD;
            }


            
            resp+=dp[check];
            swap(dp,newdp);
            swap(val,div);
        }
        int ptr = 0;
        while(ptr < ssize(val)-1 && val[ptr] <= v[n-1])
            ptr++;
        //cout << n-1 << ' ' << val[ptr] << ' ' << dp[ptr] << '\n';
        resp+=dp[ptr];
        cout << resp << '\n';
    }

    return 0;

}