T1和T3不太可做..先只放一下T2
TMD考场上没算好空间直接MLE爆零...
操作1可以归到操作3里，并且几个人的操作可以合并到一块，用线段树挺好维护的。
对于询问的话可以对每一个节点开一个桶，记录区间内前缀数量，向上合并的时候左儿子直接加，右儿子异或后再加。
发现很多节点根本用不到，动态开点即可。

#include<iostream>
#include<cstring>
#include<cstdio>
#define lson (k<<1)
#define rson ((k<<1)|1)
using namespace std;
int n, m, r, q, opt, x, y, v, ans, To, tot;
const int N = 130005;
int sum[N << 2][1 << 6], lz[N << 2];
inline int read() 
{
	int res = 0; char ch = getchar(); bool XX = false;
	for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
	for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
	return XX ? -res : res;
}
struct ju 
{
	int c[3][4];
	friend ju operator ^(const ju &a, const ju &b) 
	{
		ju c;
		memset(c.c, 0, sizeof(c.c));
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= m; ++j)
				c.c[i][j] = a.c[i][j] ^ b.c[i][j];
		return c;
	}
} to, tmp, tr[N << 2];
inline void fu(ju &x, int v) 
{
	memset(x.c, 0, sizeof(x.c));
	if (v <= n)
		for (int i = 1; i <= m; ++i)x.c[v][i] = 1;
	else 
	{
		v -= n;
		for (int i = 1; i <= n; ++i)x.c[i][v] = 1;
	}
}
inline int to1(ju x) 
{
	int res = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			res = (res << 1) | x.c[i][j];
	return res;
}
inline ju to2(int x) 
{
	ju res;
	memset(res.c, 0, sizeof(res.c));
	for (int i = n; i >= 1; --i)
		for (int j = m; j >= 1; --j)
			res.c[i][j] = (x & 1), x >>= 1;
	return res;
}
inline void upd(int k) 
{
	int tmp;
	tr[k] = tr[lson] ^ tr[rson];
	for (int i = 0; i < tot; ++i)sum[k][i] = sum[lson][i];
	tmp = to1(tr[lson]);
	for (int i = 0; i < tot; ++i)sum[k][i ^ tmp] += sum[rson][i];
}
void build(int k, int l, int r) 
{
	lz[k] = -1;
	if (l == r) 
	{
		fu(tr[k], 1); sum[k][to1(tr[k])] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(lson, l, mid); build(rson, mid + 1, r);
	upd(k);
}
inline void work(int k, int l, int r, int v) 
{
	tr[k] = to2((r - l + 1) & 1 ? v : 0);
	lz[k] = v;
	for (int i = 0; i < tot; ++i)sum[k][i] = 0;
	sum[k][v] = (r - l + 2) >> 1; sum[k][0] = (r - l + 1) - sum[k][v];
}
inline void cd(int k, int l, int r) 
{
	int mid = (l + r) >> 1;
	work(lson, l, mid, lz[k]); work(rson, mid + 1, r, lz[k]);
	lz[k] = -1;
}
void change(int k, int l, int r, int x, int y, int v) 
{
	if (x <= l && r <= y) 
	{
		work(k, l, r, v);
		return;
	}
	if (lz[k] != -1)cd(k, l, r);
	int mid = (l + r) >> 1;
	if (x <= mid)change(lson, l, mid, x, y, v);
	if (mid + 1 <= y)change(rson, mid + 1, r, x, y, v);
	upd(k);
}
int ask2(int k, int l, int r, int x, int y, int val) 
{
	if (x <= l && r <= y)return sum[k][val];
	if (lz[k] != -1)cd(k, l, r);
	int res = 0, mid = (l + r) >> 1;
	if (x <= mid)res += ask2(lson, l, mid, x, y, val);
	if (mid + 1 <= y)res += ask2(rson, mid + 1, r, x, y, val ^ to1(tr[lson]));
	return res;
}
void solve2() 
{
	ju tmp;
	build(1, 1, r);
	while (q--) 
	{
		opt = read();
		if (opt == 0) 
		{
			x = read(); y = read();
			fu(tmp, y);
			change(1, 1, r, x, x, to1(tmp));
		}
		if (opt == 1) 
		{
			x = read(); y = read();
			printf("%d\n", ask2(1, 1, r, x, y, To));
		}
		if (opt == 2) 
		{
			x = read(); y = read(); v = read();
			fu(tmp, v);
			change(1, 1, r, x, y, to1(tmp));
		}
	}
}
int main() 
{
	cin >> n >> m; tot = (1 << (n * m));
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			to.c[i][j] = read();
	To = to1(to);
	cin >> r >> q;
	solve2();
	fclose(stdin); fclose(stdout);
	return 0;
}