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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int C(long long int n, long long int r) {
  if (r > n - r) r = n - r;
  long long int ans = 1;
  for (long long int i = 1; i <= r; i++) {
    ans *= (n - r + i);
    ans /= i;
  }
  return ans;
}
long long int stringtointer(string a) { return stoi(a); }
long long int countdigit(long long int a) {
  long long int ans = 0;
  while (a) {
    a /= 10;
    ans++;
  }
  return ans;
}
long long int factorial(long long int n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long int dayofweek(long long int d, long long int m, long long int y) {
  static long long int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
  y -= m < 3;
  return (y + y / 4 - y / 100 + y / 400 + t[m - 1] + d) % 7;
}
long long int floorSqrt(long long int x) {
  if (x == 0 || x == 1) return x;
  long long int start = 1, end = x, ans;
  while (start <= end) {
    long long int mid = (start + end) / 2;
    if (mid * mid == x) return mid;
    if (mid * mid < x) {
      start = mid + 1;
      ans = mid;
    } else
      end = mid - 1;
  }
  return ans;
}
bool checkpallindrome(string s) {
  string ans = s;
  reverse(ans.begin(), ans.end());
  if (ans == s) return true;
  return false;
}
long long int kadane(long long int a[], long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
long long int calc(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  long long int res = (x % m + m) % m;
  return res;
}
void solve(long long int tc) {
  long long int n;
  cin >> n;
  cout << n << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  long long int tc = 1;
  while (t--) {
    solve(tc);
    tc++;
  }
  return 0;
}