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# Range Sum of Sorted Subarray Sums

Given the array `nums` consisting of `n` positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of `n * (n + 1) / 2` numbers.

*Return the sum of the numbers from index* `left` *to index* `right` (**indexed from 1**)*, inclusive, in the new array.* Since the answer can be a huge number return it modulo 10^9 + 7.

**Example 1:**

```
Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: 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. 
```

**Example 2:**

```
Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: 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.
```

**Example 3:**

```
Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50
```

**Constraints:**

* `1 <= nums.length <= 10^3`
* `nums.length == n`
* `1 <= nums[i] <= 100`
* `1 <= left <= right <= n * (n + 1) / 2`

```java
class Solution {
    static class Pair {
        long sum;
        int index;

        Pair(long sum, int index) {
            this.sum = sum;
            this.index = index;
        }
    }

    // The benefit of using PQ is, that instead of using N*(N+1)/2 space
    // We will only use O(N) space
    public int rangeSum(int[] nums, int n, int left, int right) {
        PriorityQueue<Pair> PQ = new PriorityQueue<>((a, b) -> (int) a.sum - (int) b.sum);
        for (int i = 0; i < n; i++)
            PQ.add(new Pair(nums[i], i + 1));
        long ans = 0, MOD = 1_000_000_007;
        for (int i = 1; i <= right; i++) {
            Pair smallSum = PQ.poll();
            if (i >= left)
                ans = (ans + smallSum.sum) % MOD;
            if (smallSum.index < n) {
                smallSum.sum = (smallSum.sum + nums[smallSum.index++]) % MOD;
                PQ.add(smallSum);
            }
        }
        return (int) ans;
    }
}
```


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