<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) {
        int i=nums.size()-n;
        int index=0;
        for(;i>0;i--)
        {
            int min=index++;
            for(int j=0;j<nums.size();j++)
            {
                if(nums[min]>nums[j])
                {
                    min=j;
                }
            }
            swap(nums[min],nums[i]);
        }
       //nums.sort(nums.begin(),nums.end());
        int *arr=new int[(n*(n+1))/2];
        int indexForarr=0;
        int sum=0;
        int modulo=(int)(pow(10,9)+7);
        for(int i=0;i<nums.size();i++)
        {
            sum=0;
            for(int j=i;j<nums.size();j++)
            {
                sum=sum+nums[j]%modulo;
                arr[indexForarr++]=sum;
            }
        }
        sum=0;
        if(left-1==0 && right==(n*(n+1))/2)
        {
            for(int i=0;i<(n*(n+1))/2;i++)
            {
                sum=(sum+arr[i])%modulo;
            }
            return sum;
        }
        for(int i=0;i<right;i++)
        {
            int min=i;
            for(int j=i+1;j<(n*(n+1))/2;j++)
            {
                if(arr[min]>arr[j])
                {
                    min=j;
                }
            }
            swap(arr[min],arr[i]);
            if(left-1<=i)
            sum=(sum+arr[i])%modulo;
        }
        return sum%modulo;
    }
};
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