// Arup Guha
// 10/12/2016
// Solution to 2016 NAQ Problem D: Brackets

import java.util.*;

public class brackets {

	public static int n;
	public static String list;
	public static int[] cumfreq;
	public static int[] minfreq;

	public static void main(String[] args) {

		Scanner stdin = new Scanner(System.in);
		list = stdin.next();
		n = list.length();

		// Store cumulative frequencies.
		cumfreq = new int[n+1];
		for (int i=1; i<=n; i++) {
			if (list.charAt(i-1) == '(')
				cumfreq[i] = cumfreq[i-1] + 1;
			else
				cumfreq[i] = cumfreq[i-1] - 1;
		}

		// Also store minimum frequencies to end. (So minfreq[i] is min of cumfreq[i..n].
		minfreq = new int[n+1];
		minfreq[n] = cumfreq[n];
		for (int i=n-1; i>=0; i--)
			minfreq[i] = Math.min(minfreq[i+1], cumfreq[i]);

		// Solve it!
		System.out.println(solve());
	}

	public static String solve() {

		// Take care of in beginning.
		int min = cumfreq[0];
		for (int i=0; i<=n; i++)
			min = Math.min(min, cumfreq[i]);
		if (min == 0 && cumfreq[n] == 0)
			return "possible";

		int curSum = cumfreq[n];

		// Try each valid starting point.
		for (int start=0; start<n; start++) {

			// Our starting point must be before here.
			if (cumfreq[start] <= -1) break;

			// Can't flip this one.
			if (cumfreq[start+1] <= 1 && list.charAt(start) == '(') continue;

			// Try each ending point.
			int curChange = 0;
			for (int end=start; end<n; end++) {

				// Update our change to the rest of the cumfreq array.
				if (list.charAt(end) == '(')
					curChange -= 2;
				else
					curChange += 2;

				// Oops, this is bad, we closed something off.
				if (curChange+cumfreq[end+1] < 0) break;

				// To be valid, there must be equal number of ( and ).
				if (2*(cumfreq[end+1]-cumfreq[start]) - curSum == 0) {

					// And every cum sum must be 0 or more.
					if (minfreq[end+1]+curChange >= 0)
						return "possible";

				}
			}
		}

		// If we get here, we can't do it.
		return "impossible";
	}
}