For an array $b$ of length $m$, define $f(b)$ as follows.

Consider a $1 \times m$ strip, where all cells initially have darkness $0$. You want to transform it into a strip where the color at the $i$-th position has darkness $b_i$. You can perform the following operation, which consists of two steps:

1. Fold the paper at any line between two cells. You may fold as many times as you like, or choose not to fold at all.
2. Choose one position to drop the black dye. The dye permeates from the top and flows down to the bottom, increasing the darkness of all cells in its path by $1$. After dropping the dye, you unfold the strip.

Let $f(b)$ be the minimum number of operations required to achieve the desired configuration. It can be proven that the goal can always be achieved in a finite number of operations.

You are given an array $a$ of length $n$. Evaluate

$$\sum_{l=1}^n\sum_{r=l}^n f(a_l a_{l+1} \ldots a_r)$$

modulo $998\,244\,353$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the array $a$.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^9$) — denoting the array $a$.

It is guaranteed that the sum of all values of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output a single integer — the required sum modulo $998\,244\,353$.

Note

In the first test case,

• $f(a_1)=f(\mathtt{0})=0$
• $f(a_1a_2)=f(\mathtt{01})=1$
• $f(a_1a_2a_3)=f(\mathtt{010})=1$
• $f(a_2)=f(\mathtt{1})=1$
• $f(a_2a_3)=f(\mathtt{10})=1$
• $f(a_3)=f(\mathtt{0})=0$

This sums up to $0+1+1+1+1+0 = 4$

In the second test case, $f(a_1a_2a_3a_4a_5a_6) = 2$. Here's one possible sequence of operations.