<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>

#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native","f16c")
static const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();
class Solution {
public:

    long long mod=1e9+7;

    long long kadane(vector<int> arr){
        long long ans=0;
        long long cs=0;

        for(int i=0;i<arr.size();i++){
            cs=(cs+arr[i]);
            ans=max(ans,cs);

            if(cs<0){
                cs=0;
            }
        }

        return ans;
    }

    int kConcatenationMaxSum(vector<int>& arr, int k) {
        long long s=accumulate(arr.begin(),arr.end(),0);

        if(k>1){
            arr.insert(arr.end(),arr.begin(),arr.end());
        }

        long long sum=kadane(arr)%mod;

        if(s>0){
            if(k>2){
                return 1LL*(1LL*sum+1LL*s*(k-2))%mod;
            }else
            return sum;
        }else{
            return sum;
        }

    }
};
48000