<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) {
        long result =
            (sumOfFirstK(nums, n, right) - sumOfFirstK(nums, n, left - 1)) %
            mod;
        // Ensure non-negative result
        return (result + mod) % mod;
    }

private:
    int mod = 1e9 + 7;
    // Helper function to count subarrays with sum <= target and their total
    // sum.
    pair<int, long long> countAndSum(vector<int>& nums, int n, int target) {
        int count = 0;
        long long currentSum = 0, totalSum = 0, windowSum = 0;
        for (int j = 0, i = 0; j < n; ++j) {
            currentSum += nums[j];
            windowSum += nums[j] * (j - i + 1);
            while (currentSum > target) {
                windowSum -= currentSum;
                currentSum -= nums[i++];
            }
            count += j - i + 1;
            totalSum += windowSum;
        }
        return {count, totalSum};
    }

    // Helper function to find the sum of the first k smallest subarray sums.
    long long sumOfFirstK(vector<int>& nums, int n, int k) {
        int minSum = *min_element(nums.begin(), nums.end());
        int maxSum = accumulate(nums.begin(), nums.end(), 0);
        int left = minSum, right = maxSum;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (countAndSum(nums, n, mid).first >= k)
                right = mid - 1;
            else
                left = mid + 1;
        }
        auto [count, sum] = countAndSum(nums, n, left);
        // There can be more subarrays with the same sum of left.
        return sum - left * (count - k);
    }
};
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