<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) {
        if(n == 1) return 0;
        vector<int> cnt(n+1,1e9);
        vector<vector<bool>> vis(n+1,vector<bool>(n+1));
        cnt[1] = 0;
        vis[1][0] = 1;
        queue<vector<int>> q;
        q.push({2,1,2}); // {curLen,how much increase,cnt}
        while(!q.empty()){
            auto it = q.front();
            q.pop();
            int x = it[0];
            int y = it[1];
            int z = it[2];
            if(x > n) continue;
            cout<<"x "<<x<<" y "<<y<<" z "<<z<<endl;
            if(vis[x][y]) continue;
            vis[x][y] = 1;
            if(cnt[x] > z) cnt[x] = z;
            q.push({x+y,y,z+1}); 
            q.push({2*x,x,z+2});
        }
        return cnt[n];
    }
};
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