<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 mod = 1e9+7;
        long long n = arr.size();
        long long sum_arr = 0;
        for(auto it : arr){
            sum_arr += it%mod;
        }
        vector<long long> pref(n);
        vector<long long> suf(n);
        long long sum = 0;
        long long max_pre = 0;
        long long max_suf = 0;
        for(int i=0; i<n; i++){
            sum += arr[i]%mod;
            pref[i] = sum%mod;
            max_pre = max(max_pre, pref[i]);
        } 
        sum = 0;
        for(int i=n-1; i>=0; i--){
            sum += arr[i]%mod;
            suf[i] = sum%mod;
            max_suf = max(max_suf, suf[i]);
        }
        sum = 0;
        long long max_sum =  0;
        for(int i=0; i<n; i++){
            sum += arr[i]%mod;
            if(sum<=0){
                sum=0;
            }
            max_sum = max(max_sum, sum); 
        }

        if(k==1){
            if(sum_arr>=0) return max_sum%mod;
            return max_sum%mod;
        }
        if(sum_arr>=0){
            
            int ans = (max_pre%mod + max_suf%mod)%mod ;
            if(k>=2){
                ans += ((k-2)*sum_arr)%mod;
            }
            return ans%mod;
        }
        else{
            int ans= max(max_pre + max_suf, max_sum)%mod;
            return ans%mod;
        }


    }
};
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