<p>You have <code>k</code> lists of sorted integers in <strong>non-decreasing&nbsp;order</strong>. Find the <b>smallest</b> range that includes at least one number from each of the <code>k</code> lists.</p>

<p>We define the range <code>[a, b]</code> is smaller than range <code>[c, d]</code> if <code>b - a &lt; d - c</code> <strong>or</strong> <code>a &lt; c</code> if <code>b - a == d - c</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
<strong>Output:</strong> [20,24]
<strong>Explanation: </strong>
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].
</pre>

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

<pre>
<strong>Input:</strong> nums = [[1,2,3],[1,2,3],[1,2,3]]
<strong>Output:</strong> [1,1]
</pre>

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

<ul>
	<li><code>nums.length == k</code></li>
	<li><code>1 &lt;= k &lt;= 3500</code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 50</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code>&nbsp;is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

class Solution {
public:
    typedef pair<int,vector<int>> pi;
    vector<int> smallestRange(vector<vector<int>>& nums) {
        priority_queue<pi,vector<pi>,greater<pi>> p;
        int row=nums.size();
        int mx=INT_MIN;
        vector<int> v1(2);
        for(int i=0;i<row;i++){
            
            v1[0]=i;
            v1[1]=0;
            mx=max(mx,nums[i][0]);
            p.push({nums[i][0],v1});
        }
        vector<int> ans(2);
        ans[0]=(p.top()).first;
        ans[1]=mx;
        while(true){
            
            vector<int> v2(2);
            v2=p.top().second;
            p.pop();
            if(v2[1]+1==nums[v2[0]].size()) break;
            else{
                v1[0]=v2[0];
                v1[1]=v2[1]+1;
                mx=max(mx,nums[v2[0]][v2[1]+1]);
                p.push({nums[v2[0]][v2[1]+1],v1});
                if(mx-p.top().first < ans[1]-ans[0]){
                    ans[0]=(p.top()).first;
                    ans[1]=mx;
                }
            }
        }
        return ans;
        
    }
};
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