<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>

class Solution {
public:
    int minSteps(int n) {
        deque<vector<int>> q;
        unordered_set<int> seen;
        seen.insert(1);
        q.push_back({0, 1, 0});
        int i = 0;
        while (1) {
            vector<int> f = q.front();
            int d = f[0], curr = f[1], p = f[2];
            //int d = q.front()[0], curr = q.front()[1], p = q.front()[2];
            q.pop_front();
            if (curr == n) return d;
            if (curr+p <= n && seen.count(curr+p+n*p) == 0) {
                q.push_back({d+1, curr+p, p});
                seen.insert(curr+p+n*p);
            }
            if (2*curr <= n && seen.count(curr+n*curr) == 0) {
                q.push_back({d+1, curr, curr});
                seen.insert(curr+n*curr);
            }
        }
        return -1;
    }
};
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