<p>Given an integer array <code>arr</code> and an integer <code>k</code>, modify the array by repeating it <code>k</code> times.</p>

<p>For example, if <code>arr = [1, 2]</code> and <code>k = 3 </code>then the modified array will be <code>[1, 2, 1, 2, 1, 2]</code>.</p>

<p>Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be <code>0</code> and its sum in that case is <code>0</code>.</p>

<p>As the answer can be very large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2], k = 3
<strong>Output:</strong> 9
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,-2,1], k = 5
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> arr = [-1,-2], k = 7
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) {
        int n=arr.size();
        int mod=1e9+7;
        int pos=0,neg=0;
        int sum=0;
        if(k==1)
        {
            int ans=0,s=0;
            for(int i=0;i<n;i++)
            {
                s+=arr[i];
                ans=max(ans,s);
                if(s<0)
                s=0;
            }
            return ans;
        }
        for(int i=0;i<n;i++)
        {
            sum+=arr[i];
            if(arr[i]>0)
            pos++;
            else if(arr[i]<0)
            neg++;
        }
        if(pos==n)
        {
            int fans=0;
            for(int i=0;i<k;i++)
            fans=(fans+sum)%mod;
            return fans;
        }
        if(neg==n)
        return 0;
        if(sum>0)
        {
            int ans=0;
            for(int i=2;i<k;i++)
            ans=(ans+sum)%mod;
            int use=ans;
            for(int i=0;i<n;i++)
            {
                use=(use+arr[i])%mod;
                ans=max(ans,use);
            }
            use=ans;
            for(int i=n-1;i>=0;i--)
            {
                use=(use+arr[i])%mod;
                ans=max(ans,use);
            }
            return ans;
        }
        else
        {
            vector<int> newArray(2*n);
            for(int i=0;i<2*n;i++)
            newArray[i]=arr[i%n];
            int ans=0,s=0;
            for(int i=0;i<2*n;i++)
            {
                s+=newArray[i];
                ans=max(ans,s);
                if(s<0)
                s=0;
            }
            return ans;
        }
    }
};
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