<p>You are given the array <code>nums</code> consisting of <code>n</code> positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of <code>n * (n + 1) / 2</code> numbers.</p>

<p><em>Return the sum of the numbers from index </em><code>left</code><em> to index </em><code>right</code> (<strong>indexed from 1</strong>)<em>, inclusive, in the new array. </em>Since the answer can be a huge number return it modulo <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 5
<strong>Output:</strong> 13 
<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 3, right = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 10
<strong>Output:</strong> 50
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= left &lt;= right &lt;= n * (n + 1) / 2</code></li>
</ul>

class Solution {
public:
    int MOD = 1e9 + 7;
    int getSum(int l, int r, vector<int>& pre){
        if(l == 0)
            return pre[r];
        return pre[r] - pre[l-1];
    }
    void display(vector<int>& vec){
        for(int& i : vec)
            cout<<i<<" ";
        cout<<endl;
    }
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        // good exercise on prefix sums
        vector<int>prenum(n, 0);
        prenum[0] = nums[0];
        for(int i=1;i<n;i++){
            prenum[i] = nums[i] + prenum[i-1];
        }
        // cout<<"PF of nums"<<endl;
        // display(prenum);
        vector<int>subs;
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                // subarray i to j
                subs.push_back(getSum(i, j, prenum));
            }
        }
        sort(subs.begin(), subs.end());
        // cout<<"Sorted subarray sums "<<endl;
        // display(subs);
        vector<int>pres(subs.size(), 0);
        pres[0] = subs[0];
        for(int i=1;i<subs.size();i++){
            pres[i] = (subs[i] + pres[i-1]) % MOD;
        }
        // cout<<"PF of subs"<<endl;
        // display(pres);
        return getSum(left-1, right-1, pres);
    }
};
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