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;

/*-------------------------------TEMPLATE ------ START-------------------*/
//defines
#define openin freopen("res/input.txt","r",stdin)
#define openout freopen("res/output.txt","w",stdout)
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pi pair<int,int>
#define vi vector<int>
#define vvi vector<vi>
#define ll long long
#define int long long
#define repr(i,x,y) for (decltype(x) i=x;i>=y;i--)
#define rep(i,x,y) for (decltype(x) i=x;i<=y;i++)
#define all(c) (c).begin(),(c).end()
#define allr(c) (c).rbegin(),(c).rend()
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define endl "\n"

#define mod 1000000007
//#define M_PI 3.141592653589793
#define eps 1e-9

/* Print pair */
template <typename T, typename S>
ostream& operator << (ostream& os, const pair<T, S>& v) {
    os << "(";
    os << v.first << "," << v.second << ")";
    return os;
}
/* Print vector */
template <typename T>
ostream& operator << (ostream& os, const vector<T>& v) {
    os << "[";
    int sz = v.size();
    for (int i = 0; i < sz; ++i) {
        os << v[i];
        if (i != sz - 1)os << ",";
    }
    os << "]\n";
    return os;
}
/* Print list */
template <typename T>
ostream& operator << (ostream& os, const list<T>& v) {
    os << "[";
    int sz = v.size();
    int i = 0;
    for (auto x : v) {
        os << x;
        if (i != sz - 1)os << ",";
        ++i;
    }
    os << "]\n";
    return os;
}
/* Print set */
template <typename T>
ostream& operator << (ostream& os, const set<T>& v) {
    T last = *v.rbegin();
    os << "[";
    for (auto it : v) {
        os << it;
        if (it != last) os << ",";
    }
    os << "]\n";
    return os;
}
/* Print Map */
template <typename T, typename S>
ostream& operator << (ostream& os, const map<T, S>& v) {
    for (auto it : v) {
        os << it.first << " : " << it.second << "\n";
    }
    return os;
}
int power(int a, int b, int m = mod)
{
    int res = 1;
    while (b)
    {
        if (b % 2) {
            res = (res%m * a%m) % m;
        }
        b /= 2;
        a = (a%m * a%m) % m;
    }
    return res;
}

/* String to int */
uint64_t toint(const string& s) {
    istringstream src(s);
    uint64_t v;
    src >> v;
    return v;
}

/*Int to string */
string tostring(uint64_t v) {
    ostringstream src;
    src << v;
    return src.str();
}

int get_mid(int a, int b) {
    return a + (b - a) / 2;
}

int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

int ceil_a_by_b(int a, int b) {
    return (a + b - 1) / b;
}

int _lcm(int a, int b) {
    return (a * b) / gcd(a, b);
}

/*-------------------------------TEMPLATE ------ END-------------------*/
//SOLUTION

const int N = 1000;

void solve() {
    int n;
    cin >> n;

    auto re = [&]() {
        vector<string> ip(n);
        rep(i, 0, n - 1) cin >> ip[i];
        vvi v(n, vi(n));
        rep(i, 0, n - 1)
            rep(j, 0, n - 1)
            v[i][j] = ip[i][j] == '1';
        return v;
    };

    vvi a = re();
    vvi b = re();

    //cout << a << b;
    vvi d(n, vi(n)); //diff matrix

    rep(i, 0, n - 1) {
        rep(j, 0, n - 1) {
            d[i][j] = a[i][j] ^ b[i][j];
        }
    }

    auto f = [n](vvi v, bool inv) {

        auto row = [&](int i) {
            rep(j, 0, n - 1) {
                v[i][j] ^= 1;
            }
        };
        auto col = [&](int j) {
            rep(i, 0, n - 1) {
                v[i][j] ^= 1;
            }
        };

        if (inv) {
            row(0);
        }
        if(v[0][0])
            col(0);

        rep(i, 1, n-1) if (v[i][0]) row(i);
        rep(i, 1, n-1) if (v[0][i]) col(i);

        rep(i, 0, n - 1) {
            rep(j, 0, n - 1) {
                if (v[i][j]) return false;
            }
        }
        return true;
    };

    cout << (f(d, true) or f(d, false) ? "YES" : "NO");
}


int32_t main()
{
#ifndef ONLINE_JUDGE 
    openin;
#endif // !ONLINE_JUDGE

    
    fast;
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
        cout << endl;
    }

    return 0;

}