<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return the maximum sum of a <strong>non-empty</strong> subsequence of that array such that for every two <strong>consecutive</strong> integers in the subsequence, <code>nums[i]</code> and <code>nums[j]</code>, where <code>i &lt; j</code>, the condition <code>j - i &lt;= k</code> is satisfied.</p>

<p>A <em>subsequence</em> of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,2,-10,5,20], k = 2
<strong>Output:</strong> 37
<b>Explanation:</b> The subsequence is [10, 2, 5, 20].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,-2,-3], k = 1
<strong>Output:</strong> -1
<b>Explanation:</b> The subsequence must be non-empty, so we choose the largest number.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,-2,-10,-5,20], k = 2
<strong>Output:</strong> 23
<b>Explanation:</b> The subsequence is [10, -2, -5, 20].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
    priority_queue<pair<int, int>> pq;
    int maxt = 0, i = 0;

  public:
    Solution() {
        pq.push({0, -1});
    }
    int constrainedSubsetSum(const vector<int> &nums, const int k) {
        int maxte = INT_MIN;
        for (const int &i : nums)
            maxte = max(i, maxte);
        if (maxte <= 0)
            return maxte;

        for (const int &n : nums) {
            while (!pq.empty() && i - pq.top().second > k)
                pq.pop();

            const int a = max(n, n + pq.top().first);

            if (pq.top().first <= a)
                pq = {};
                
            maxt = max(maxt, a);
            pq.push({a, i});
            i++;
        }

        return maxt;
    }
};
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