You are given two binary square matrices a and b of size n × n. A matrix is called binary if each of its elements is equal to 0 or 1. You can do the following operations on the matrix a arbitrary number of times (0 or more): 

  * vertical xor. You choose the number j (1 ≤ j ≤ n) and for all i (1 ≤ i ≤ n) do the following: a_{i, j} := a_{i, j} ⊕ 1 (⊕ — is the operation [xor](https://en.wikipedia.org/wiki/Exclusive_or) (exclusive or)). 
  * horizontal xor. You choose the number i (1 ≤ i ≤ n) and for all j (1 ≤ j ≤ n) do the following: a_{i, j} := a_{i, j} ⊕ 1. 



Note that the elements of the a matrix change after each operation.

For example, if n=3 and the matrix a is: $$$ \begin{pmatrix} 1 & 1 & 0 \\\ 0 & 0 & 1 \\\ 1 & 1 & 0 \end{pmatrix} $$$ Then the following sequence of operations shows an example of transformations: 

  * vertical xor, j=1. $$$ a= \begin{pmatrix} 0 & 1 & 0 \\\ 1 & 0 & 1 \\\ 0 & 1 & 0 \end{pmatrix} $$$ 
  * horizontal xor, i=2. $$$ a= \begin{pmatrix} 0 & 1 & 0 \\\ 0 & 1 & 0 \\\ 0 & 1 & 0 \end{pmatrix} $$$ 
  * vertical xor, j=2. $$$ a= \begin{pmatrix} 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \end{pmatrix} $$$ 



Check if there is a sequence of operations such that the matrix a becomes equal to the matrix b.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 1000) — the size of the matrices.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix a.

An empty line follows.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix b.

It is guaranteed that the sum of n over all test cases does not exceed 1000.

Output

For each test case, output on a separate line: 

  * "YES", there is such a sequence of operations that the matrix a becomes equal to the matrix b; 
  * "NO" otherwise. 



You can output "YES" and "NO" in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

Example

Input


3
3
110
001
110

000
000
000
3
101
010
101

010
101
010
2
01
11

10
10


Output


YES
YES
NO

Note

The first test case is explained in the statements.

In the second test case, the following sequence of operations is suitable: 

  * horizontal xor, i=1; 
  * horizontal xor, i=2; 
  * horizontal xor, i=3; 



It can be proved that there is no sequence of operations in the third test case so that the matrix a becomes equal to the matrix b.
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef pair<int ,int> pii;
typedef pair<ll, ll> pll;
typedef double rl;

typedef pii int2;
typedef tuple<int, int, int>           int3;
typedef tuple<int, int, int, int>      int4;
typedef tuple<int, int, int, int, int> int5;

typedef pll ll2;
typedef tuple<ll, ll, ll>         ll3;
typedef tuple<ll, ll, ll, ll>     ll4;
typedef tuple<ll, ll, ll, ll, ll> ll5;

#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
#define g4(x) get<4>(x)

typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vll;
typedef vector<pll> vll_ll;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
typedef vector<vll> vvll;
typedef vector<vll_ll> vvll_ll;

#define pb push_back
#define mp make_pair
#define eb emplace_back
#define xx first
#define yy second

#define FOR(i,N) for(int i=0; i<N; i++)

#define popc __builtin_popcountll
// #define popc __builtin_popcount

template<typename A>
string to_string(A* ptr)
{
    stringstream ss; ss << "0x" << std::setw(16) << std::setfill ('0') << std::hex << (uint64_t)(uintptr_t)ptr;
    return ss.str();
}

string to_string(char c)
{
    return ((string)"'"+c)+ "'";
}

template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#if !ONLINE_JUDGE && 1
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define dbg(...) 42
#endif

const int MX=1009;
int n;
string a[MX];
string b[MX];

int A[MX][MX];
int R[MX];
int C[MX];

bool solve()
{
    FOR(i, n)
    {
        FOR(j, n)
        {
            int aa=a[i][j]-'0';
            int bb=b[i][j]-'0';
            A[i][j]=aa^bb;
        }
    }

    for(int r0=0; r0<=1; r0++)
    {
        R[0]=r0;
        for(int j=0; j<n; j++) C[j]=A[0][j]^R[0];

        for(int i=1; i<n; i++)
        {
            R[i]=C[0]^A[i][0];
        }

        bool ok=true;
        FOR(i, n) FOR(j, n) if(A[i][j]!=(R[i]^C[j])) ok=false;

        if(ok) return true;
    }

    return false;
}

int main()
{
#if !ONLINE_JUDGE && 1
    freopen("F.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
#endif
#if 0
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
#endif

    int TT;
#if 1
    cin >> TT;
#else
    TT=1;
#endif
    for(int ttt=0; ttt<TT; ttt++)
    {
        cin >> n;
        FOR(i, n) cin >> a[i];
        FOR(i, n) cin >> b[i];

        cout << (solve()?"YES":"NO") << '\n';
    }

    return 0;
}
