http://acm.hdu.edu.cn/showproblem.php?pid=5755
题意：一个N*M的矩阵，改变一个格子，本身+2，四周+1.同时mod 3;问操作多少次，矩阵变为全0.输出次数和具体位置
由于影响是相互的，所以增广矩阵的系数a[t][t+1] 或者是 a[t+1][t]均可；只需注意往结果中添加位置时，x[i]表示要操作x[i]次，所以位置要重复添加x[i]次；
高斯消元时间复杂度为O（N3M3）;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define pb push_back
#define A first
#define B second
#define MK make_pair
#define inf 0x3f3f3f3f
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define lowbit(x) (x&(-x))
#define clear0 (0xFFFFFFFE)
#define mod 3
#define K(x) ((x)*(x))
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
T x = ,f = ;char ch = getchar();
while(ch <'' || ch >''){ if(ch == '-') f = -;ch=getchar(); }
while(ch >= '' && ch <= ''){ x = x* + ch - '';ch = getchar(); }
m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
if(a>) out(a/);
putchar(a%+'');
}
inline ll gcd(ll a,ll b){ return b == ? a: gcd(b,a%b); }
inline ll lcm(ll a,ll b){ return a/gcd(a,b)*b; }
const int maxn = ;
int a[maxn][maxn];
int x[maxn<<]; void init(int n,int m)
{
rep0(i,,n) rep0(j,,m){
int t = i*m + j;
a[t][t] = ;
if(i > ) a[t - m][t] = ;
if(i < n-) a[t + m][t] = ;
if(j > ) a[t-][t] = ;
if(j < m-) a[t+][t] = ;
}
} int equ, var;
int Gauss()
{
int mx, row, col;
for(row = , col = ; row < equ && col < var; row++, col++){
mx = row;
for(int i = row+;i < equ;i++){
if(abs(a[i][col]) > abs(a[mx][col])) mx = i;
}
if(a[mx][col] == ) {
row--;
continue;
}
if(mx != row){
for(int j = col;j < var + ;j++)
swap(a[row][j], a[mx][j]);
}
for(int i = row+;i < equ;i++){
if(a[i][col]){
int LCM = lcm(abs(a[i][col]), abs(a[row][col]));
int ta = LCM/abs(a[i][col]), tb = LCM/abs(a[row][col]);
if(a[i][col] * a[row][col] < ) tb = -tb;
for(int j = col; j < var+;j++)
a[i][j] = ((a[i][j]*ta - a[row][j]*tb)%mod+mod)%mod;
}
}
} for(int i = var-; i >= ; i--){
if(a[i][i] == ) continue;
int tmp = a[i][var];
for(int j = i+;j < var;j++){
if(a[i][j]){
tmp -= a[i][j]*x[j];
tmp = (tmp%mod + mod)% mod;
}
}
x[i] = (tmp*a[i][i])% mod;
}
return ;
}
vector<int> vec;
int main()
{
//freopen("data.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T, kase = ;
scanf("%d",&T);
while(T--){
int n, m, t;
read2(n,m);
equ = n*m, var = n*m;
MS0(a);
rep0(i,,n) rep0(j,,m) read1(t), a[i*m+j][var] = (mod-t)%mod;
init(n,m); Gauss();
vec.clear();
rep0(i,,var) while(x[i]--) vec.pb(i);
out(vec.size());puts("");
rep0(i,,vec.size()) printf("%d %d\n",vec[i]/m + , vec[i]%m + );
}
return ;
}