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 <unordered_map>
#define DEBUG
#undef DEBUG
using namespace std;
typedef int ll;
typedef pair<int, int> pt;

const int MOD = 998'244'353;

void solve_std() {
	int n;
	cin >> n;
	vector<int> a(n);
	vector<int> mini(n);
	for (auto& ai : a)cin >> ai;

	ll sum = 0;
	ll all_cnt = 0;

	for (int k = n - 1; k >= 0; k--) {
		mini[k] = a[k];
		for (int i = k + 1; i < n; i++) {
			if (a[k] > mini[i]) {
				int cnt = ceil(1. * a[k] / mini[i]);
				mini[i] = min((int)(a[k] / cnt), mini[i]);
				all_cnt = (all_cnt + cnt - 1) % MOD;
			}
			else {
				mini[i] = a[k];
			}
			//printf("%d ", mini[i]);
		}
		//printf("\n");
		all_cnt %= MOD;
		sum = (sum + all_cnt) % MOD;
	}

	cout << sum << endl;
}

void solve1() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (auto& ai : a)cin >> ai;

	ll sum = 0;
	ll all_cnt = 0;

	vector<pt> miniMap;
	vector<pt> newMap;

	for (int k = n - 1; k >= 0; k--) {
		newMap.clear();
		int cur = a[k];
		newMap.emplace_back(cur, 1);
		for (auto mi : miniMap) {
			int mini = mi.first;
			int occur = mi.second;
			if (cur <= mini) {
				newMap[0].second += occur;
			}
			else {
				int cnt = ceil(1. * cur / mini);
				mini = min((int)(cur / cnt), mini);
				all_cnt = (all_cnt + (cnt - 1LL) * occur) % MOD;
				if (newMap.back().first == mini)
					newMap.back().second += occur;
				else
					newMap.emplace_back(mini, occur);
			}
		}

		miniMap.swap(newMap);
		sum = (sum + all_cnt) % MOD;
	}

	cout << sum << endl;
}

void solve() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (auto& ai : a)cin >> ai;

	ll sum = 0;
	ll all_cnt = 0;
	
	unordered_map<int,int> miniMap;
	unordered_map<int, int> newMap;

	for (int k = n - 1; k >= 0; k--) {
		newMap.clear();
		int cur = a[k];
		newMap.emplace(cur, 1);
		for (auto mi : miniMap) {
			int mini = mi.first;
			int occur = mi.second;
			if (cur <= mini) {
				newMap[cur] += occur;
			}
			else {
				int cnt = (cur + mini - 1)/ mini;
				//mini = min(cur / cnt, mini);
				int tmp = (cnt - 1) * occur % MOD;
				all_cnt = (all_cnt + tmp) % MOD;
				newMap[cur / cnt] += occur;
			}
		}

		miniMap.swap(newMap);
		sum = (sum + all_cnt) % MOD;
	}

	cout << sum << '\n';
}

int main() {
#ifdef DEBUG    
	freopen("input.txt", "r", stdin);
#endif
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	int t = 1;
	cin >> t;
	while (t--) {
		//solve_std();
		solve();
	}
	return 0;
}