<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 rangeSum(vector<int>& nums, int n, int left, int right) {
        vector<vector<int>> vvi(n);
        int i, j;
        int mod = 1e9 + 7;
        vvi[0].push_back(nums[0]);
        for(i = 1; i < n; i++) {
            nums[i] = nums[i] + nums[i - 1];
            vvi[0].push_back(nums[i]);
        }
        for(i = 0; i < n - 1; i++) {
            for(j = i + 1; j < n; j++) {
                vvi[i+1].push_back(nums[j] - nums[i]);
            }
        }

        priority_queue<pair<int,pair<int,int>>, vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int>>>> pq;

        for(i = 0; i < vvi.size(); i++) {
            pq.push({vvi[i][0], {i, 0}});
        }
        long long int lsum = 0, rsum = 0;
        int val, ind = 0;
        pair<int, int> row;
        while(!pq.empty()) {
            auto ele = pq.top();
            pq.pop();
            val = ele.first;
            ind++;
            row = ele.second;
            if (vvi[row.first].size() > row.second + 1) {
                pq.push({vvi[row.first][row.second + 1], {row.first, row.second + 1}});
            }
            // cout<<"val = "<<val<<" row = "<<row.first<<" "<<row.second + 1<<endl;
            if (ind >= left && ind <= right) {
                lsum = (lsum + val) % mod;
            }
            // cout<<"lsum = "<<lsum<<" ind = "<<ind<<endl;
            
            if (ind > right) {
                break;
            }
        }
        return lsum % mod;
    }
};
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