human
I found an interesting problem on https://codeforces.com/problemset/problem/1151/E:
The Kingdom of Kremland is a tree (a connected undirected graph without cycles) consisting of $n$ vertices. Each vertex $i$ has its own value $a_i$. All vertices are connected in series by edges. Formally, for every $1 \leq i < n$ there is an edge between the vertices of $i$ and $i+1$.

Denote the function $f(l, r)$, which takes two integers $l$ and $r$ ($l \leq r$):

    We leave in the tree only vertices whose values ​​range from $l$ to $r$.    The value of the function will be the number of connected components in the new graph. 

Your task is to calculate the following sum: $$\sum_{l=1}^{n} \sum_{r=l}^{n} f(l, r) $$

-----Input-----

The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of vertices in the tree.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the values of the vertices.

-----Output-----

Print one number — the answer to the problem.

-----Examples-----
Input
3
2 1 3

Output
7
Input
4
2 1 1 3

Output
11
Input
10
1 5 2 5 5 3 10 6 5 1

Output
104

-----Note-----

In the first example, the function values ​​will be as follows:   $f(1, 1)=1$ (there is only a vertex with the number $2$, which forms one component)  $f(1, 2)=1$ (there are vertices $1$ and $2$ that form one component)  $f(1, 3)=1$ (all vertices remain, one component is obtained)  $f(2, 2)=1$ (only vertex number $1$)  $f(2, 3)=2$ (there are vertices $1$ and $3$ that form two components)  $f(3, 3)=1$ (only vertex $3$)  Totally out $7$.

In the second example, the function values ​​will be as follows:   $f(1, 1)=1$  $f(1, 2)=1$  $f(1, 3)=1$  $f(1, 4)=1$  $f(2, 2)=1$  $f(2, 3)=2$  $f(2, 4)=2$  $f(3, 3)=1$  $f(3, 4)=1$  $f(4, 4)=0$ (there is no vertex left, so the number of components is $0$)  Totally out $11$.

I tried it in Python, but could not do it. Can you solve it?

=====
Can you provide a Python code to print one number — the answer to the problem?
assistant
#Here is Python code to print one number — the answer to the problem.

# AC
import sys

class Main:
    def __init__(self):
        self.buff = None
        self.index = 0

    def __next__(self):
        if self.buff is None or self.index == len(self.buff):
            self.buff = sys.stdin.readline().split()
            self.index = 0
        val = self.buff[self.index]
        self.index += 1
        return val

    def next_int(self):
        return int(next(self))

    def solve(self):
        n = self.next_int()
        x = [self.next_int() for _ in range(0, n)]
        ans = 0
        for i in range(0, n):
            ans += x[i] * (n - x[i] + 1)
            if i > 0:
                ans -= min(x[i], x[i - 1]) * (n - max(x[i], x[i - 1]) + 1)
        print(ans)

def __starting_point():
    Main().solve()

__starting_point()