<p>An attendance record for a student can be represented as a string where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:</p>

<ul>
	<li><code>&#39;A&#39;</code>: Absent.</li>
	<li><code>&#39;L&#39;</code>: Late.</li>
	<li><code>&#39;P&#39;</code>: Present.</li>
</ul>

<p>Any student is eligible for an attendance award if they meet <strong>both</strong> of the following criteria:</p>

<ul>
	<li>The student was absent (<code>&#39;A&#39;</code>) for <strong>strictly</strong> fewer than 2 days <strong>total</strong>.</li>
	<li>The student was <strong>never</strong> late (<code>&#39;L&#39;</code>) for 3 or more <strong>consecutive</strong> days.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the <strong>number</strong> of possible attendance records of length</em> <code>n</code><em> that make a student eligible for an attendance award. The answer may be very large, so return it <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 8
<strong>Explanation:</strong> There are 8 records with length 2 that are eligible for an award:
&quot;PP&quot;, &quot;AP&quot;, &quot;PA&quot;, &quot;LP&quot;, &quot;PL&quot;, &quot;AL&quot;, &quot;LA&quot;, &quot;LL&quot;
Only &quot;AA&quot; is not eligible because there are 2 absences (there need to be fewer than 2).
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 3
</pre>

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

<pre>
<strong>Input:</strong> n = 10101
<strong>Output:</strong> 183236316
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:

    int mod=1e9+7;
    int checkRecord(int n) {
        vector<vector<long long>>dp(n+3,vector<long long>(3,0));
        dp[1][0]=1;
        dp[1][1]=1;
        dp[1][2]=0;
        for(int i=2;i<=n;i++){
            dp[i][0]=(dp[i-1][0]+dp[i-1][1]+dp[i-1][2])%mod;
            dp[i][0]%=mod;
            //late
            dp[i][1]+=dp[i-1][0];
            dp[i][2]+=dp[i-1][1];
            dp[i][1]%=mod;
            dp[i][2]%=mod;
        }
        long long ans=0;
        ans+=dp[n][0];
        ans%=mod;
        ans+=dp[n][1];
        ans%=mod;
        ans+=dp[n][2];
        ans%=mod;
        cout<<ans<<endl;
        for(int i=1;i<=n;i++){
            int left=i-1;
            int right=n-i;
            long long lefts=1;
            long long rights=1;
            if(left!=0){
                lefts=0;
                lefts+=dp[left][0];
        lefts%=mod;
        lefts+=dp[left][1];
        lefts%=mod;
        lefts+=dp[left][2];
        lefts%=mod;
            }
            if(right!=0){
                rights=0;
                rights+=dp[right][0];
        rights%=mod;
        rights+=dp[right][1];
        rights%=mod;
        rights+=dp[right][2];
        rights%=mod;
            }
            // if(!(left==0&&right==0)){
                ans+=((rights)*lefts)%mod;
                ans%=mod;
            // }
        }
        return ans;
    }
};
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