<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 ans = 0;
        int mod = 1e9 + 7;
        int ans1 = 0;
        int n = arr.size();
        vector<int>	pref(n+1);
        int maxi=0;
        for (long long int i = 1; i <= n; ++i) {
            pref[i]=arr[i-1];
            pref[i] += pref[i - 1];
            maxi=max(pref[i],maxi);
        }

        if (k == 1) {
            long long int sum = 0, maxi = 0;
            for (long long int i = 0; i < n; ++i) {
                sum += arr[i];
                if (sum <= 0) {
                    sum = 0;
                }
                maxi = max(maxi, sum);
            }
            return maxi;
        }
        vector<int>copy=arr;
        for(int i:arr){
            copy.push_back(i);
        }
        int sum=0;
        for(int i=0; i<copy.size();++i){
            sum+=copy[i];
            if(sum<0){
                sum=0;
            }
            maxi=fmax(maxi,sum);
        }
        if(pref[n]>0){
            pref[n]%=mod;
            long long val = (((pref[n]*1ll*(k-2))%mod)+mod)%mod;
            return (maxi+val)%mod;
        }
        else{
            return maxi;
        }
    }
};
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