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> 
#include <ext/pb_ds/assoc_container.hpp> 
using namespace std;
using namespace __gnu_pbds;
// Policy based data structure 
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 
#define int long long
#define ll long long 
#define pb push_back
#define pii pair<int,int>
#define rep(i,st,en) for(int i=st;i<en;i++)
#define vi vector<int>
#define vii vector<pii>
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define eb emplace_back
#define yes cout<<"YES"<<endl; return;
#define no cout<<"NO"<<endl; return;
#define flus fflush(stdout);
#define gin(x) cerr<<#x<<" : "<<x<<" ";
#define fin cerr<<endl;
#define F first
#define S second
#define np next_permutation
#define inf 1e18
#define mod 1000000007
#define N 200009
#define PI 3.14159265358979323846
#define minpq priority_queue <ll, vector<ll>, greater<ll>>
void solve(){
   int n;
   cin >> n;
   string s[2];
   for(int i=0;i<2;i++) cin >> s[i];
   for(int i=1;i<s[0].length();i++)
   {
       if(s[0][i]=='1' && s[1][i]=='1' && (s[0][i-1]=='1' || s[1][i-1]=='1'))
       {
           cout << "NO" << endl;
           return;
       }
   }
   cout << "YES" << endl;
}
signed main(){
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

 