#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (auto &it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
inline T Bit(T x, long long i) {
  return (x >> i) & 1;
}
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long N = 2e5 + 3;
void solve(long long tc) {
  long long n;
  cin >> n;
  string s[2];
  for (long long i = 0; i < 2; i++) cin >> s[i];
  long long a[2][n];
  memset(a, 0, sizeof(a));
  a[0][0] = 1;
  if (s[1][0] == '0') a[1][0] = 1;
  for (long long i = 1; i < n; i++) {
    if (a[0][i - 1] || a[1][i - 1]) {
      if (s[0][i] == '0') a[0][i] = 1;
      if (s[1][i] == '0') a[1][i] = 1;
    }
  }
  if (a[1][n - 1]) {
    cout << "YES" << '\n';
    return;
  };
  {
    cout << "NO" << '\n';
    return;
  };
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

### Prompt

Generate a cpp solution to 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 <class T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (auto &it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
inline T Bit(T x, long long i) {
  return (x >> i) & 1;
}
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long N = 2e5 + 3;
void solve(long long tc) {
  long long n;
  cin >> n;
  string s[2];
  for (long long i = 0; i < 2; i++) cin >> s[i];
  long long a[2][n];
  memset(a, 0, sizeof(a));
  a[0][0] = 1;
  if (s[1][0] == '0') a[1][0] = 1;
  for (long long i = 1; i < n; i++) {
    if (a[0][i - 1] || a[1][i - 1]) {
      if (s[0][i] == '0') a[0][i] = 1;
      if (s[1][i] == '0') a[1][i] = 1;
    }
  }
  if (a[1][n - 1]) {
    cout << "YES" << '\n';
    return;
  };
  {
    cout << "NO" << '\n';
    return;
  };
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

```