#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline istream &operator>>(istream &s, vector<T> &v) {
  for (T &t : v) {
    s >> t;
  }
  return s;
}
template <typename T>
inline ostream &operator<<(ostream &s, const vector<T> &v) {
  for (int i = 0; i < int(v.size()); ++i) {
    s << (" " + !i) << v[i];
  }
  return s;
}
template <typename T>
inline T fromString(const string &s) {
  T res;
  istringstream iss(s);
  iss >> res;
  return res;
};
template <typename T>
inline string toString(const T &a) {
  ostringstream oss;
  oss << a;
  return oss.str();
};
const long long mod = 1000000007;
const long long INF = 1e18;
int main() {
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    char s[2][n];
    for (int i = (0); i < (2); i++) {
      for (int j = (0); j < (n); j++) {
        cin >> s[i][j];
      }
    }
    long long f = 0;
    for (int i = (0); i < (1); i++) {
      for (int j = (0); j < (n); j++) {
        if (s[i][j] == '1' && s[i + 1][j] == '1') {
          f = 1;
        }
      }
    }
    if (f) {
      cout << "NO"
           << "\n";
    } else {
      cout << "YES"
           << "\n";
    }
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline istream &operator>>(istream &s, vector<T> &v) {
  for (T &t : v) {
    s >> t;
  }
  return s;
}
template <typename T>
inline ostream &operator<<(ostream &s, const vector<T> &v) {
  for (int i = 0; i < int(v.size()); ++i) {
    s << (" " + !i) << v[i];
  }
  return s;
}
template <typename T>
inline T fromString(const string &s) {
  T res;
  istringstream iss(s);
  iss >> res;
  return res;
};
template <typename T>
inline string toString(const T &a) {
  ostringstream oss;
  oss << a;
  return oss.str();
};
const long long mod = 1000000007;
const long long INF = 1e18;
int main() {
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    char s[2][n];
    for (int i = (0); i < (2); i++) {
      for (int j = (0); j < (n); j++) {
        cin >> s[i][j];
      }
    }
    long long f = 0;
    for (int i = (0); i < (1); i++) {
      for (int j = (0); j < (n); j++) {
        if (s[i][j] == '1' && s[i + 1][j] == '1') {
          f = 1;
        }
      }
    }
    if (f) {
      cout << "NO"
           << "\n";
    } else {
      cout << "YES"
           << "\n";
    }
  }
}

```