Problem: You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes. 

Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it. 

2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.

Please look at pictures below for better understanding.

<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.

<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4. 

You have to answer t independent test cases.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^{4}) — the number of test cases.

Each of the next t lines contains a single integer n (1 ≤ n ≤ 10^{9}).

Output

For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.

Example

Input


2
2
1


Output


2
1

Note

In the first test case, there are the following 2 ways to fill the area:

<image>

In the second test case, there is a unique way to fill the area:

<image>

Solution (in PYTHON3):

Solution:

def main():
    # from collections import defaultdict
    # from collections import Counter
    # import math
    # from fractions import Fraction
    # import heapq
    # from bisect import bisect_left
    import sys
    input = sys.stdin.buffer.readline
    #mod=998244353
    def GCD(x, y):
        while (y):
            x, y = y, x % y
        return x
    def lcm(a,b):
        return (a*b)//GCD(a,b)
    # Modified merge for inversion
    def merge(a,start,mid,end,d):
        p=start
        q=mid+1
        arr=[0]*(end-start+1)
        k=0
        coun=0
        for i in range(start,end+1):
            if p > mid:
                arr[k]=a[q]
                q+=1
                k+=1
            elif q > end:
                arr[k]=a[p]
                d[a[p]] += coun
                k+=1
                p+=1
            elif a[p] < a[q]:
                arr[k]=a[p]
                d[a[p]]+=coun
                k+=1
                p+=1
            else:
                arr[k]=a[q]
                k+=1
                q+=1
                coun+=1

        for i in range(k):
            a[start]=arr[i]
            start+=1
    def mergesort(a, start, end,d):
        if start<end:
            mid=(start+end)//2
            mergesort(a,start,mid,d)
            mergesort(a,mid+1,end,d)
            merge(a,start,mid,end,d)
    # inversion return dic for every index inversion count
    def inversion(l):
        d={}
        for i in l:
            d[i]=0
        mergesort(l,0,len(l)-1,d)
        return d
    def binary_search_lower(A,low,high,k):
        if high-low!=1:
            mid=(low+high)//2
            x = A[mid]
            if x>=k:
                return binary_search_lower(A,low,mid,k)
            else:
                return binary_search_lower(A,mid,high,k)
        if A[low]>=k:
            return low
        else:
            return high
    # Function to find power
    def binary_expo(x,n):
        if n==0:
            return 1
        elif n%2==0:
            return binary_expo(x*x,n//2)
        else:
            return x*binary_expo(x*x,(n-1)//2)
    # Function to find power with given modulus
    def modu_expo(x,n,m):
        if n==0:
            return 1
        elif n%2==0:
            return modu_expo((x*x)%m,n//2,m)
        else:
            return (x*modu_expo((x*x)%m,(n-1)//2,m))%m
    # Function for modular multiplicative inverse via using extended_GCD_Euclid's
    def eEuclid(c, m):
        if m == 0:
            global d
            global x
            global y
            d = c
            x = 1
            y = 0
        else:
            eEuclid(m, c % m)
            temp = x
            x = y
            y = temp - (c // m) * y
    def mod_Inv(c, m):
        eEuclid(c, m)
        return (x % m + m) % m
    # Function for nCr%p using fermat (p is prime)
    def ncr(n, r, p):
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % p
            den = (den * (i + 1)) % p
        return (num * pow(den,
                          p - 2, p)) % p
    #*************************************************************
    for _ in range(int(input())):
        n=int(input())
        print(n)

if __name__ == "__main__":
    main()