Monocarp is playing a computer game. Now he wants to complete the first level of this game.

A level is a rectangular grid of 2 rows and n columns. Monocarp controls a character, which starts in cell (1, 1) — at the intersection of the 1-st row and the 1-st column.

Monocarp's character can move from one cell to another in one step if the cells are adjacent by side and/or corner. Formally, it is possible to move from cell (x_1, y_1) to cell (x_2, y_2) in one step if |x_1 - x_2| ≤ 1 and |y_1 - y_2| ≤ 1. Obviously, it is prohibited to go outside the grid.

There are traps in some cells. If Monocarp's character finds himself in such a cell, he dies, and the game ends.

To complete a level, Monocarp's character should reach cell (2, n) — at the intersection of row 2 and column n.

Help Monocarp determine if it is possible to complete the level.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow. Each test case consists of three lines.

The first line contains a single integer n (3 ≤ n ≤ 100) — the number of columns.

The next two lines describe the level. The i-th of these lines describes the i-th line of the level — the line consists of the characters '0' and '1'. The character '0' corresponds to a safe cell, the character '1' corresponds to a trap cell.

Additional constraint on the input: cells (1, 1) and (2, n) are safe.

Output

For each test case, output YES if it is possible to complete the level, and NO otherwise.

Example

Input


4
3
000
000
4
0011
1100
4
0111
1110
6
010101
101010


Output


YES
YES
NO
YES

Note

Consider the example from the statement.

In the first test case, one of the possible paths is (1, 1) → (2, 2) → (2, 3).

In the second test case, one of the possible paths is (1, 1) → (1, 2) → (2, 3) → (2, 4).

In the fourth test case, one of the possible paths is (1, 1) → (2, 2) → (1, 3) → (2, 4) → (1, 5) → (2, 6).
1
9
000000010
000000000
 1
9
000000010
000100000
 1
60
000000000000000000000000000000000000000000000000000000000010
000000000000000000000000000000000000000000000000000000000010

YES
 YES
 NO

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
void write(vector<long long> &v) {
  for (auto i : v) cout << i;
  cout << "\n";
}
void read(vector<long long> &v) {
  for (auto &i : v) cin >> i;
}
void solve() {
  long long n;
  cin >> n;
  char a[2][n];
  for (long long(i) = (0); (i) < (2); i++) {
    for (long long(j) = (0); (j) < (n); j++) {
      cin >> a[i][j];
    }
  }
  for (long long(j) = (0); (j) < (n - 1); j++) {
    if (a[0][j] == '0' || a[1][j] == '0' || a[0][j + 1] == '0' ||
        a[1][j + 1] == '0')
      continue;
    else {
      cout << "NO" << endl;
      return;
    }
  }
  if (a[1][n - 1] == '0')
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
