Polycarp has a string s[1 ... n] of length n consisting of decimal digits. Polycarp performs the following operation with the string s no more than once (i.e. he can perform operation 0 or 1 time): 

  * Polycarp selects two numbers i and j (1 ≤ i ≤ j ≤ n) and removes characters from the s string at the positions i, i+1, i+2, …, j (i.e. removes substring s[i ... j]). More formally, Polycarp turns the string s into the string s_1 s_2 … s_{i-1} s_{j+1} s_{j+2} … s_{n}. 



For example, the string s = "20192020" Polycarp can turn into strings: 

  * "2020" (in this case (i, j)=(3, 6) or (i, j)=(1, 4)); 
  * "2019220" (in this case (i, j)=(6, 6)); 
  * "020" (in this case (i, j)=(1, 5)); 
  * other operations are also possible, only a few of them are listed above. 



Polycarp likes the string "2020" very much, so he is wondering if it is possible to turn the string s into a string "2020" in no more than one operation? Note that you can perform zero operations.

Input

The first line contains a positive integer t (1 ≤ t ≤ 1000 ) — number of test cases in the test. Then t test cases follow.

The first line of each test case contains an integer n (4 ≤ n ≤ 200) — length of the string s. The next line contains a string s of length n consisting of decimal digits. It is allowed that the string s starts with digit 0.

Output

For each test case, output on a separate line: 

  * "YES" if Polycarp can turn the string s into a string "2020" in no more than one operation (i.e. he can perform 0 or 1 operation); 
  * "NO" otherwise. 



You may print every letter of "YES" and "NO" in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


6
8
20192020
8
22019020
4
2020
5
20002
6
729040
6
200200


Output


YES
YES
YES
NO
NO
NO

Note

In the first test case, Polycarp could choose i=3 and j=6.

In the second test case, Polycarp could choose i=2 and j=5.

In the third test case, Polycarp did not perform any operations with the string.
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 

"""**************************MAIN*****************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        s=sin()
        x="2020"
        ans="NO"
        if s[:1]+s[n-3:]==x or s[:2]+s[n-2:]==x or s[:3]+s[n-1:]==x or s[:4]+s[n:]==x or s[n-4:]==x :
            ans="YES"
        print(ans)





"""***********************************************"""

######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()