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>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vii = vector<pii>;
struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
template <typename T>
void ndarray(vector<T> &vec, int len) {
  vec.resize(len);
}
template <typename T, typename... Args>
void ndarray(vector<T> &vec, int len, Args... args) {
  vec.resize(len);
  for (auto &v : vec) ndarray(v, args...);
}
template <typename V, typename T>
void ndfill(V &x, const T &val) {
  x = val;
}
template <typename V, typename T>
void ndfill(vector<V> &vec, const T &val) {
  for (auto &v : vec) ndfill(v, val);
}
template <typename T>
bool chmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool chmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) {
  return make_pair(l.first + r.first, l.second + r.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) {
  return make_pair(l.first - r.first, l.second - r.second);
}
template <typename T>
vector<T> srtunq(vector<T> vec) {
  sort(vec.begin(), vec.end()),
      vec.erase(unique(vec.begin(), vec.end()), vec.end());
  return vec;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << '[';
  for (auto v : vec) os << v << ',';
  os << ']';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &vec) {
  os << "deq[";
  for (auto v : vec) os << v << ',';
  os << ']';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &vec) {
  os << '{';
  for (auto v : vec) os << v << ',';
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &vec) {
  os << '{';
  for (auto v : vec) os << v << ',';
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &vec) {
  os << '{';
  for (auto v : vec) os << v << ',';
  os << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) {
  os << '{';
  for (auto v : vec) os << v << ',';
  os << '}';
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &pa) {
  os << '(' << pa.first << ',' << pa.second << ')';
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const map<TK, TV> &mp) {
  os << '{';
  for (auto v : mp) os << v.first << "=>" << v.second << ',';
  os << '}';
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) {
  os << '{';
  for (auto v : mp) os << v.first << "=>" << v.second << ',';
  os << '}';
  return os;
}
constexpr ll MOD = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> a(2, vector<int>(n));
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < n; j++) {
      char c;
      cin >> c;
      a[i][j] = c - '0';
    }
  }
  vector<int> b(n);
  for (int i = 0; i < n; i++) {
    if (a[0][i] == 1 && a[1][i] == 1) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
}
int main() {
  fast_ios_;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
