#include <bits/stdc++.h>
using namespace std;
long long secondmax(long long a, long long b, long long c) {
  long long ar[3];
  ar[0] = a, ar[1] = b, ar[2] = c;
  sort(ar, ar + 3);
  return ar[1];
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first)
    return a.first > b.first;
  else
    return a.second > b.second;
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
long long pow1(long long base, long long x) {
  long long ans = 1;
  for (long long i = 1; i <= x; i++) ans *= base;
  return ans;
}
void debug(string s) { cout << s << "\n"; }
void debug(long long s) { cout << s << "\n"; }
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    if (s1[n - 1] != '0' && s2[n - 1] != '0') {
      cout << "NO\n";
      continue;
    }
    long long c = 0;
    for (long long i = 0; i < n; i++) {
      if (s1[i] == '0' || s2[i] == '0') c++;
    }
    if (c == n)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long secondmax(long long a, long long b, long long c) {
  long long ar[3];
  ar[0] = a, ar[1] = b, ar[2] = c;
  sort(ar, ar + 3);
  return ar[1];
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first)
    return a.first > b.first;
  else
    return a.second > b.second;
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
long long pow1(long long base, long long x) {
  long long ans = 1;
  for (long long i = 1; i <= x; i++) ans *= base;
  return ans;
}
void debug(string s) { cout << s << "\n"; }
void debug(long long s) { cout << s << "\n"; }
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    if (s1[n - 1] != '0' && s2[n - 1] != '0') {
      cout << "NO\n";
      continue;
    }
    long long c = 0;
    for (long long i = 0; i < n; i++) {
      if (s1[i] == '0' || s2[i] == '0') c++;
    }
    if (c == n)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```