human
Alica and Bob are playing a game.

Initially they have a binary string s consisting of only characters 0 and 1.

Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible: 

  1. delete s_1 and s_2: 1011001 → 11001.

  2. delete s_2 and s_3: 1011001 → 11001.

  3. delete s_4 and s_5: 1011001 → 10101.

  4. delete s_6 and s_7: 1011001 → 10110. 


If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.

Input

First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.

Output

For each test case print answer in the single line.

If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.

Example

Input

3
01
1111
0011

Output

DA
NET
NET

Note

In the first test case after Alice's move string s become empty and Bob can not make any move.

In the second test case Alice can not make any move initially.

In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move.
Solve the task in PYTHON3.
 ... Can you provide a Python code to determine if Alice can win?
assistant
#Here is Python code to determine if Alice can win.

## necessary imports
import sys
input = sys.stdin.readline
#from math import ceil, floor, factorial;

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if b == 0:
        return a
    return gcd(b, a % b);

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return int(res) 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return...