<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>

#define F first
#define S second
class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& a) {
    int k;k=a.size();
    // priority_queue<pair<int,int> >pq;
    set<pair<int,pair<int,int> > >st;
    vector<int>v;
    for(int i=0;i<k;i++) st.insert(make_pair(a[i][0],make_pair(i,0)));
    map<int,int>mp;
    while(1){
        auto it=st.begin();
        int l=(*it).F;
        it=st.end();
        it--;
        int r=(*it).F;
        // cout<<l<<" "<<r<<endl;
        if(!mp.count(r-l)) mp[r-l]=l;
        else mp[r-l]=min(mp[r-l],l);
        it=st.begin();
        int i=(*it).S.F,j=(*it).S.S;
        st.erase(it);
        j++;
        if(j==a[i].size()) break;
        st.insert(make_pair(a[i][j],make_pair(i,j)));
    }
    auto it=mp.begin();
    v.push_back(it->S);v.push_back(it->F+it->S);
    return v;
    }
};
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