<p>Given a balanced parentheses string <code>s</code>, return <em>the <strong>score</strong> of the string</em>.</p>

<p>The <strong>score</strong> of a balanced parentheses string is based on the following rule:</p>

<ul>
	<li><code>&quot;()&quot;</code> has score <code>1</code>.</li>
	<li><code>AB</code> has score <code>A + B</code>, where <code>A</code> and <code>B</code> are balanced parentheses strings.</li>
	<li><code>(A)</code> has score <code>2 * A</code>, where <code>A</code> is a balanced parentheses string.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;()&quot;
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(())&quot;
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;()()&quot;
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 50</code></li>
	<li><code>s</code> consists of only <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>
	<li><code>s</code> is a balanced parentheses string.</li>
</ul>

class Solution:
    def scoreOfParentheses(self, s: str) -> int:
        stack=[0]
        for i in s:
            if i=="(":
                stack.append(0)
            else:
                value= max(2*stack.pop(),1)
                stack[-1]+= value
        return stack.pop()
16400