#include <bits/stdc++.h>
using namespace std;
int a[2005][2005];
int a2[2005][2005];
int n;
const int T = 3;
bool ok[2005];

int I[2005];
const int mod = 1e9 + 7;
typedef pair<int,int> pii;
mt19937 rnd(0);

int AI[2005];
int AI2[2005];
void test() {
    for(int i = 1;i <= n;i++) I[i] = rnd()%(mod - 1) + 1;
    for(int i = 1;i <= n;i++) { //[i,1]
        AI[i] = 0;
        for(int j = 1;j <= n;j++) { //A[i][j]
            AI[i] = (AI[i] + 1LL * a[i][j] * I[j]) % mod;
        }
    }
    for(int i = 1;i <= n;i++) {
        AI2[i] = 0;
        for(int j = 1;j <= n;j++) {
            AI2[i] = (AI2[i] + 1LL * a2[i][j] * AI[j]) % mod;
        }
        AI2[i] = (I[i] - AI2[i] + mod) % mod;
    }
    for(int i = 1;i <= n;i++) {
        if(AI2[i] != 0) ok[i] = 1; ///第i行
    }
}
vector<int> get_row() {
    memset(ok,0,sizeof(ok)) ;
    for(int i = 1;i <= T;i++) {
        test() ;
    }
    vector<int> ans;
    for(int i = 1;i <= n;i++) {
        if(ok[i]) ans.push_back(i) ;
    }
    return ans;
}
vector<int> row , col;
int id(int i,int j) {
    return (i - 1) * col.size() + j ;
}
pair<int,int> to(int k) {
    int c = (k - 1) % col.size() + 1;
    int r = (k - c) / col.size() + 1;
    // printf("RC %d %d\n",r,c);
    return (pair<int,int>){row[r - 1] , col[c - 1]} ;
}
int mat[505][505] , b[505];
int sol[505];
int fpow(int a,int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a %mod;
        a = 1LL * a * a %mod; b>>= 1;
    }
    return ans;
}
void solve(int n,int cnt) {
    vector<int> id(n + 1);
    
    for(int i = 1;i < n;i++) {
        int cur = -1;
        for(int j = i;j <= cnt;j++) {
            if(mat[j][i]) {cur = j ;break ;}
        }
        assert(cur != -1) ;
        for(int j = 1;j <= n;j++) swap(mat[i][j] , mat[cur][j]) ;
        swap(b[i] , b[cur]) ;
        int rr = fpow(mat[i][i] , mod - 2) ;
        for(int j = i + 1;j <= cnt;j++) {
            int f = (mod - 1LL * mat[j][i] * rr%mod) % mod;
            for(int k = i;k <= n;k++) mat[j][k] = (mat[j][k] + 1LL * mat[i][k] * f) % mod;
            b[j] = (b[j] + 1LL * b[i] * f) % mod;
        }
    }
    // for(int i = 1;i <= n;i++) {
    //     for(int j = 1;j <= n;j++) printf("%d ",mat[i][j]) ;
    //     printf(": %d\n",b[i]) ;
    // }
    for(int i = n;i >= 1;i--) {
        int sum = b[i];
        for(int j = i + 1;j <= n;j++) {
            sum  = (sum - 1LL * sol[j] * mat[i][j]%mod + mod)  % mod;
        }
        sol[i] = 1LL * sum * fpow(mat[i][i] , mod - 2) % mod;
        // printf("%d %d %d\n",i,sol[id[i]] , id[i]);
    }
    return ;
}
int main() {
    ios::sync_with_stdio(false) ; cin.tie(0) ; cout.tie(0) ;
    cin >> n;
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= n;j++) cin >> a[i][j];
    }
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= n;j++) cin >> a2[i][j];
    }
    row = get_row() ;
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j < i;j++) swap(a[i][j] , a[j][i]) , swap(a2[i][j] , a2[j][i]) ;
    }
    col = get_row() ;
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j < i;j++) swap(a[i][j] , a[j][i]) , swap(a2[i][j] , a2[j][i]) ;
    }
    int cnt = 0;
    // printf("R %d %d\n",row.size() , col.size()) ;
    for(int i = 0;i < col.size();i++) {
        for(int j = 1;j <= min(n , 2*(int)row.size());j++) { ///A的第j行，和A'的col[i]列
            int cc = 0; ++cnt ;
            if(j == col[i]) b[cnt] = 1;
            else b[cnt] = 0;
            for(int k = 1;k <= n;k++) {
                if(cc < col.size() && k == row[cc]) { //第k列
                    mat[cnt][id(cc + 1 , i + 1)] = a[j][k] ;
                    ++cc ;
                }
                b[cnt] = (b[cnt] - 1LL * a[j][k] * a2[k][col[i]]%mod + mod) % mod;
            }
        }
    }
    // for(int i = 1;i <= cnt;i++) {
    //     for(int j = 1;j <= cnt;j++) printf("%d ",mat[i][j]) ;
    //     printf(": %d\n",b[i]) ;
    // }
    solve(col.size() * row.size() , cnt) ;
    vector<pair<pii,int> > ans;
    for(int i = 1;i <= cnt;i++) {
        if(sol[i]) {
            // printf("I %d %d\n",i,sol[i]) ;
            pii t = to(i) ;
            ans.push_back({t , (a2[t.first][t.second] + sol[i]) % mod}) ;
        }
    }
    cout << ans.size() << '\n' ;
    sort(ans.begin() , ans.end()) ;
    for(auto [x,y] : ans) {
        cout << x.first <<' ' << x.second <<' ' << y << '\n' ;
    }
    return 0;
}