<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>

const int MODULO = 1000000007;
class Solution {
    vector<int> stk;
    void merge(vector<int>&ar, int sl, int el, int sr, int er) {
        int k = 0, i = sl, j = sr;
        while ((i <= el) && (j <= er)) {
            if (ar[i] < ar[j]) {
                stk[k] = ar[i];
                k++, i++;
            } else {
                stk[k] = ar[j];
                k++, j++;
            }
        }

        while (i <= el) {
            stk[k] = ar[i];
            k++, i++;
        }
        while (j <= er) {
            stk[k] = ar[j];
            k++, j++;
        }

        for (int x=0, y = sl; x < k; x++, y++) {
            ar[y] = stk[x];
        }
    }

    void mergeSort(vector<int>&ar, int s, int e) {
        if (s < e) {
            int mid = s + ((e - s)/2);
            mergeSort(ar, s, mid);
            mergeSort(ar, mid + 1, e);
            merge(ar, s, mid, mid + 1, e);
        }
    }

public:
    int rangeSum(vector<int>& nums, int n, int left, int right) {

        vector<int> sums(0);
        for (int i = 0; i < nums.size(); i++) {
            int sum = 0;
            for (int j = i; j < nums.size(); j++) {
                sum += nums[j] % MODULO;
                // cout << sum << " ";
                sums.push_back(sum);
            }
            // cout << endl;
        }

        stk.resize(sums.size(), 0);


        // cout << "calling merge sort: ";

        mergeSort(sums, 0, sums.size() - 1);
        // for (auto s : sums) {
        //     cout << s << " ";
        // }
        // cout << endl;
        int sum = 0;
        for (int i = left - 1; i < right; i++) {
            sum += sums[i];
            sum %= MODULO;
        }
        return sum;
    }
};
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