CF1438E.Yurii Can Do Everything

Yurii is sure he can do everything. Can he solve this task, though?

He has an array $a$ consisting of $n$ positive integers. Let's call a subarray $a[l...r]$ good if the following conditions are simultaneously satisfied:

• $l+1 \leq r-1$, i. e. the subarray has length at least $3$;
• $(a_l \oplus a_r) = (a_{l+1}+a_{l+2}+\ldots+a_{r-2}+a_{r-1})$, where $\oplus$ denotes the bitwise XOR operation.

In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.

Yurii wants to calculate the total number of good subarrays. What is it equal to?

An array $c$ is a subarray of an array $d$ if $c$ can be obtained from $d$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

尤里确信自己无所不能。但他能解决这个题目吗？

他有一个由 $n$ 个正整数组成的数组 $a$。我们称子数组 $a[l...r]$ 是“好的”，当且仅当同时满足以下两个条件：

• $l+1 \leq r-1$，即该子数组长度至少为 $3$；
• $(a_l \oplus a_r) = (a_{l+1}+a_{l+2}+\ldots+a_{r-2}+a_{r-1})$，其中 $\oplus$ 表示按位异或运算。

换言之，一个子数组是“好的”，当且仅当其两端元素的按位异或值等于其余所有元素之和。

尤里希望计算出“好的”子数组的总数。这个总数是多少？

数组 $c$ 是数组 $d$ 的一个子数组，当且仅当 $c$ 可通过从 $d$ 的开头删除若干（可能为零个或全部）元素，并从 $d$ 的末尾删除若干（可能为零个或全部）元素而得到。

The first line contains a single integer $n$ ($3 \leq n \leq 2\cdot 10^5$) — the length of $a$.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1 \leq a_i \lt 2^{30}$) — elements of $a$.

第一行包含一个整数 $n$（$3 \leq n \leq 2\cdot 10^5$）—— 数组 $a$ 的长度。

第二行包含 $n$ 个整数 $a_1,a_2,\ldots,a_n$（$1 \leq a_i \lt 2^{30}$）—— 数组 $a$ 的元素。

Output a single integer — the number of good subarrays.

输出一个整数——好子数组的个数。

There are $6$ good subarrays in the example:

• $[3,1,2]$ (twice) because $(3 \oplus 2) = 1$;
• $[1,2,3]$ (twice) because $(1 \oplus 3) = 2$;
• $[2,3,1]$ because $(2 \oplus 1) = 3$;
• $[3,1,2,3,1,2,3,15]$ because $(3 \oplus 15) = (1+2+3+1+2+3)$.

示例中有 $6$ 个“好”子数组：

• $[3,1,2]$（出现两次），因为 $(3 \oplus 2) = 1$；
• $[1,2,3]$（出现两次），因为 $(1 \oplus 3) = 2$；
• $[2,3,1]$，因为 $(2 \oplus 1) = 3$；
• $[3,1,2,3,1,2,3,15]$，因为 $(3 \oplus 15) = (1+2+3+1+2+3)$。