Here, N = min(k, n).

K = min(k, mn)

where m, n is the size of two arrays and k is the k in the problem.

```
class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int,int>> result;
if (nums1.empty() || nums2.empty() || k <= 0)
return result;
auto comp = [&nums1, &nums2](pair<int, int> a, pair<int, int> b) {
return nums1[a.first] + nums2[a.second] > nums1[b.first] + nums2[b.second];};
priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(comp)> min_heap(comp);
min_heap.emplace(0, 0);
while(k-- > 0 && min_heap.size())
{
auto idx_pair = min_heap.top(); min_heap.pop();
result.emplace_back(nums1[idx_pair.first], nums2[idx_pair.second]);
if (idx_pair.first + 1 < nums1.size())
min_heap.emplace(idx_pair.first + 1, idx_pair.second);
if (idx_pair.first == 0 && idx_pair.second + 1 < nums2.size())
min_heap.emplace(idx_pair.first, idx_pair.second + 1);
}
return result;
}
};
```