<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

const static auto fast = []
{ 
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

class Solution {
public:
    class Node {
        public:
        int n;
        int c;
        int p;
        Node() : n(0), c(0), p(0) {}
        Node(int n, int c, int p) : n(n), c(c), p(p) {}
    };
    int minSteps(int n) {
        deque<Node> dq;
        dq.push_back(Node(1, 0, 0));
        for (;;) {
            Node x = dq.front();
            dq.pop_front();
            if (x.n > 1000)
                continue;
            if (x.n == n)
                return x.c;
            if (x.p > 0)
                dq.push_back(Node(x.n + x.p, x.c + 1, x.p));
            if (x.p != x.n)
                dq.push_back(Node(x.n, x.c + 1, x.n));
        }
    }
};
107479