<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:
    const int MOD = 1e9 + 7;
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        int ans = 0;
        vector<int> prefixSum = vector<int>(n+1);
        vector<int> prefixPrefixSum = vector<int>(n+1);
        for(int i = 0; i < n; i++) {
            prefixSum[i+1] = prefixSum[i] + nums[i];
        }
        for(int i = 0; i < n; i++) {
            prefixPrefixSum[i+1] = prefixPrefixSum[i] + prefixSum[i+1];
        }

        return (getSum(prefixSum, prefixPrefixSum, n, right) - getSum(prefixSum, prefixPrefixSum, n, left-1)) % MOD;
    }

    int getSum(vector<int>& prefixSum, vector<int>& prefixPrefixSum, int n, int k) {
        int num = getKth(prefixSum, n, k);
        int sum = 0;
        int cnt = 0;
        for(int i = 0, j = 1; i < n; i++) {
            while(j <= n && prefixSum[j] - prefixSum[i] < num) {
                j++;
            }
            j--;
            // add sum of subarrays that are less than num
            sum += prefixPrefixSum[j] - prefixPrefixSum[i] - prefixSum[i] * (j-i);
            sum %= MOD;
            cnt += j-i;
        }
        // add sum of subarrays that are equal to num
        sum += num * (k - cnt);
        return sum;
    }

    int getKth(vector<int>& prefixSum, int n, int k) {
        int l = 0, r = prefixSum[n];
        while(l < r) {
            int mid = l + (r-l)/2;
            int cnt = getCount(prefixSum, n, mid);
            if(cnt < k) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        return l;
    }

    int getCount(vector<int>& prefixSum, int n, int x) {
        int cnt = 0;
        for(int i = 0, j = 1; i < n; i++) {
            while(j <= n && prefixSum[j] - prefixSum[i] <= x) {
                j++;
            }
            j--;
            cnt += j-i;
        }
        return cnt;
    }
};
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