#include <fstream>
#include <deque>
#include <algorithm>
#include <string.h>
#pragma warning(disable:4996)
using namespace std;

class InParser
{
private:
	FILE* fin;
	char* buff;
	int sp;
	char read()
	{
		++sp;
		if (sp == 4096)
		{
			sp = 0;
			fread(buff, 1, 4096, fin);
		}
		return buff[sp];
	}
public:
	InParser(const char* nume)
	{
		sp = 4095;
		buff = new char[4096];
		fin = fopen(nume, "r");
	}
	InParser& operator >> (int& n)
	{
		char c;
		while (!isdigit(c = read()));
		n = c - '0';
		while (isdigit(c = read()))
			n = n * 10 + c - '0';
		return *this;
	}
};

InParser fin("struti.in");
ofstream fout("struti.out");

deque <int> minq;
deque <int> maxq;

const int NMAX = 1e3 + 2;

int a[NMAX][NMAX], cnt, n, m, min_crt, p;
int mini[NMAX][NMAX], maxi[NMAX][NMAX];


void solve(int x, int y)
{
	for (int i = 1; i <= n; ++i)
	{
		minq.clear();
		maxq.clear();
		for (int j = 1; j <= m; ++j)
		{
			while (!minq.empty() && minq.front() < j - x + 1)
				minq.pop_front();

			while (!minq.empty() && a[i][minq.back()] >= a[i][j])
				minq.pop_back();

			minq.push_back(j);

			mini[i][j] = a[i][minq.front()];

			while (!maxq.empty() && maxq.front() < j - x + 1)
				maxq.pop_front();

			while (!maxq.empty() && a[i][maxq.back()] <= a[i][j])
				maxq.pop_back();

			maxq.push_back(j);
			maxi[i][j] = a[i][maxq.front()];
		}
	}
	for (int j = 1; j <= m; ++j)
	{
		minq.clear();
		maxq.clear();
		for (int i = 1; i <= n; ++i)
		{
			while (!minq.empty() && minq.front() < i - y + 1)
				minq.pop_front();

			while (!minq.empty() && mini[minq.back()][j] >= mini[i][j])
				minq.pop_back();

			minq.push_back(i);

			while (!maxq.empty() && maxq.front() < i - y + 1)
				maxq.pop_front();

			while (!maxq.empty() && maxi[maxq.back()][j] <= maxi[i][j])
				maxq.pop_back();

			maxq.push_back(i);

			if (i >= y && j >= x)
			{
				int s = maxi[maxq.front()][j] - mini[minq.front()][j];
				if (s < min_crt)
				{
					min_crt = s;
					cnt = 1;
				}
				else if (s == min_crt) ++cnt;
			}
		}
	}
}

int x, y;

int main()
{
	fin >> n >> m >> p;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			fin >> a[i][j];

	while (p--)
	{
		fin >> x >> y;
		min_crt = 10000;
		cnt = 0;
		solve(x, y);
		if (x != y) solve(y, x);
		fout << min_crt << " " << cnt << "\n";
	}
	return 0;
}