Codeforces Round 800 (Div. 2)
B
Paranoid String
Codeforces
1200
原题
← AC →
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Let's call a binary string $T$ of length $m$ indexed from $1$ to $m$ paranoid if we can obtain a string of length $1$ by performing the following two kinds of operations $m-1$ times in any order :

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  • Select any substring of $T$ that is equal to 01, and then replace it with 1.
  • Select any substring of $T$ that is equal to 10, and then replace it with 0.For example, if $T = $ 001, we can select the substring $[T_2T_3]$ and perform the first operation. So we obtain $T = $ 01.
翻译中…

You are given a binary string $S$ of length $n$ indexed from $1$ to $n$. Find the number of pairs of integers $(l, r)$ $1 \le l \le r \le n$ such that $S[l \ldots r]$ (the substring of $S$ from $l$ to $r$) is a paranoid string.

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Input

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The first line contains an integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of test cases follows.

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

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The second line of each test case contains a binary string $S$ of $n$ characters $S_1S_2 \ldots S_n$. ($S_i = $ 0 or $S_i = $ 1 for each $1 \le i \le n$)

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It is guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

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Output

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For each test case, output the number of pairs of integers $(l, r)$ $1 \le l \le r \le n$ such that $S[l \ldots r]$ (the substring of $S$ from $l$ to $r$) is a paranoid string.

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Note

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In the first sample, $S$ already has length $1$ and doesn't need any operations.

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In the second sample, all substrings of $S$ are paranoid. For the entire string, it's enough to perform the first operation.

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In the third sample, all substrings of $S$ are paranoid except $[S_2S_3]$, because we can't perform any operations on it, and $[S_1S_2S_3]$ (the entire string).

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样例

Input #1
5
1
1
2
01
3
100
4
1001
5
11111
Output #1
1
3
4
8
5
InputCopy OutputCopy
题目信息
题号
1694B
时间限制
1 s
内存限制
256 MB
标签
构造
贪心

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