Problem: Vus the Cossack has n real numbers a_i. It is known that the sum of all numbers is equal to 0. He wants to choose a sequence b the size of which is n such that the sum of all numbers is 0 and each b_i is either ⌊ a_i ⌋ or ⌈ a_i ⌉. In other words, b_i equals a_i rounded up or down. It is not necessary to round to the nearest integer.

For example, if a = [4.58413, 1.22491, -2.10517, -3.70387], then b can be equal, for example, to [4, 2, -2, -4]. 

Note that if a_i is an integer, then there is no difference between ⌊ a_i ⌋ and ⌈ a_i ⌉, b_i will always be equal to a_i.

Help Vus the Cossack find such sequence!

Input

The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of numbers.

Each of the next n lines contains one real number a_i (|a_i| < 10^5). It is guaranteed that each a_i has exactly 5 digits after the decimal point. It is guaranteed that the sum of all the numbers is equal to 0.

Output

In each of the next n lines, print one integer b_i. For each i, |a_i-b_i|<1 must be met.

If there are multiple answers, print any.

Examples

Input


4
4.58413
1.22491
-2.10517
-3.70387


Output


4
2
-2
-4


Input


5
-6.32509
3.30066
-0.93878
2.00000
1.96321


Output


-6
3
-1
2
2

Note

The first example is explained in the legend.

In the second example, we can round the first and fifth numbers up, and the second and third numbers down. We can round the fourth number neither up, nor down.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;

public class round571d2d {

	public static void main(String args[]) {
		FastScanner in = new FastScanner(System.in);
		int n = in.nextInt();
		int[] nums = new int[n];
		boolean[] canChange = new boolean[n];
		for (int i = 0; i < n; i++) {
			String next = in.next();
			String[] parts = next.split("\\.");
			
			if (next.charAt(0) != '-') {
				if (parts[1].equals("00000")) {
					canChange[i] = false;
					nums[i] = Integer.parseInt(parts[0]);
				}
				else {
					canChange[i] = true;
					nums[i] = Integer.parseInt(parts[0]);
				}
			}
			else {
				if (parts[1].equals("00000"))  {
					canChange[i] = false;
					nums[i] = Integer.parseInt(parts[0]);
				}
				else {
					canChange[i] = true;
					nums[i] = Integer.parseInt(parts[0]) - 1;
				}
			}
		}
		long sum = 0;
		for (int i = 0; i < n; i++) {
			sum += nums[i];
		}
		long count = -1* sum;
		int index = 0;
		while (count >0) {
			while (!canChange[index]) {
				
				index++;
			}
			nums[index]++;
			index++;
			count--;
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n; i++) {
			sb.append(nums[i] + "\n");
		}
		System.out.println(sb);
	}

	// ======================================================================================
	// =============================== Reference Code =======================================
	// ======================================================================================

	static int greatestDivisor(int n) {
		int limit = (int) Math.sqrt(n);
		int max = 1;
		for (int i = 2; i <= limit; i++) {
			if (n % i == 0) {
				max = Integer.max(max, i);
				max = Integer.max(max, n / i);
			}
		}
		return max;
	}

	// Method to return all primes smaller than or equal to 
	// n using Sieve of Eratosthenes 
	static boolean[] sieveOfEratosthenes(int n) {
		// Create a boolean array "prime[0..n]" and initialize 
		// all entries it as true. A value in prime[i] will 
		// finally be false if i is Not a prime, else true. 
		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i <= n; i++)
			prime[i] = true;

		prime[0] = false;
		prime[1] = false;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime 
			if (prime[p] == true) {
				// Update all multiples of p 
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		return prime;
	}

	// Binary search for number greater than or equal to target
	// returns -1 if number not found
	private static int bin_gteq(int[] a, int key) {
		int low = 0;
		int high = a.length;
		int max_limit = high;
		while (low < high) {
			int mid = low + (high - low) / 2;
			if (a[mid] < key) {
				low = mid + 1;
			} else
				high = mid;
		}

		return high == max_limit ? -1 : high;
	}

	public static int gcd(int a, int b) {
		if (a == 0)
			return b;

		return gcd(b % a, a);
	}

	static class Tuple<X, Y> {
		public final X x;
		public final Y y;

		public Tuple(X x, Y y) {
			this.x = x;
			this.y = y;
		}

		public String toString() {
			return "(" + x + "," + y + ")";
		}
	}

	static class Tuple3<X, Y, Z> {
		public final X x;
		public final Y y;
		public final Z z;

		public Tuple3(X x, Y y, Z z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public String toString() {
			return "(" + x + "," + y + "," + z + ")";
		}
	}

	static Tuple3<Integer, Integer, Integer> gcdExtended(int a, int b, int x, int y) {
		// Base Case 
		if (a == 0) {
			x = 0;
			y = 1;
			return new Tuple3(0, 1, b);
		}

		int x1 = 1, y1 = 1; // To store results of recursive call 
		Tuple3<Integer, Integer, Integer> tuple = gcdExtended(b % a, a, x1, y1);
		int gcd = tuple.z;
		x1 = tuple.x;
		y1 = tuple.y;

		// Update x and y using results of recursive 
		// call 
		x = y1 - (b / a) * x1;
		y = x1;

		return new Tuple3(x, y, gcd);
	}

	// Returns modulo inverse of a  
	// with respect to m using extended 
	// Euclid Algorithm. Refer below post for details: 
	// https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/ 
	static int inv(int a, int m) {
		int m0 = m, t, q;
		int x0 = 0, x1 = 1;

		if (m == 1)
			return 0;

		// Apply extended Euclid Algorithm 
		while (a > 1) {
			// q is quotient 
			q = a / m;

			t = m;

			// m is remainder now, process 
			// same as euclid's algo 
			m = a % m;
			a = t;

			t = x0;

			x0 = x1 - q * x0;

			x1 = t;
		}

		// Make x1 positive 
		if (x1 < 0)
			x1 += m0;

		return x1;
	}

	// k is size of num[] and rem[]. 
	// Returns the smallest number 
	// x such that: 
	// x % num[0] = rem[0], 
	// x % num[1] = rem[1], 
	// .................. 
	// x % num[k-2] = rem[k-1] 
	// Assumption: Numbers in num[] are pairwise  
	// coprime (gcd for every pair is 1) 
	static int findMinX(int num[], int rem[], int k) {
		// Compute product of all numbers 
		int prod = 1;
		for (int i = 0; i < k; i++)
			prod *= num[i];

		// Initialize result 
		int result = 0;

		// Apply above formula 
		for (int i = 0; i < k; i++) {
			int pp = prod / num[i];
			result += rem[i] * inv(pp, num[i]) * pp;
		}

		return result % prod;
	}

	/**
	 * Source: Matt Fontaine
	 */
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int chars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		int read() {
			if (chars == -1)
				throw new InputMismatchException();
			if (curChar >= chars) {
				curChar = 0;
				try {
					chars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (chars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}

}