You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.

Input

The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.

Output

The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.

If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.

Examples

Input

? + ? - ? + ? + ? = 42


Output

Possible
9 + 13 - 39 + 28 + 31 = 42


Input

? - ? = 1


Output

Impossible


Input

? = 1000000


Output

Possible
1000000 = 1000000
var rebus = readline().split(' ').map(String);

var count = {plus: 1, minus: 0}; // number of + and - in rebus (add +1 to 'plus' because of the very first '?' is + by default)

for (var i = 0; i < rebus.length; i++) {
	if (rebus[i] == '+') {
		count.plus += 1;
	} else if (rebus[i] == '-') {
		count.minus += 1;
	}
	if (rebus[i] == '=') {
		n = Number(rebus[i + 1]);
	}
}

// used for look_for parameter
var plus = '+'; 
var minus = '-';

// if possible get rebus with this function
var getRebus = function(increase, ones, look_for) {
	var sum = 0;
	if (increase == count.plus) {
		sum = ones + n;
	} else {
		sum = ones - n;
	}
	
	var diff = sum - increase;
	var a = Math.floor(diff / (n - 1)); // a --> change a number of ?'s to n
	var b = diff % (n - 1) + 1; // b --> change another ? to b
	var counter = 0; // used to count how many ?'s are replaced with n (count a)

	// if diff is 0, I want every ? to be 1
	if (diff === 0) {
		rebus[0] = 1;
	// if I want 'n' to appear 0 times, the first ? must be b (ONLY do this when increasing +?'s)
	} else if (a === 0 && increase == count.plus) {
		rebus[0] = b;
	} else {
		// ONLY when increasing +?'s
		if (increase == count.plus) {
			rebus[0] = n;
			counter = 1;
		// ONLY when increasing -?'s
		} else {
			rebus[0] = 1;
		}
		for (var i = 1; i < rebus.length; i++)  {
			// I only want to replace ? with n a times, the next replacement should be b, the rest should be 1
			if (counter == a) {
				if (rebus[i] == look_for) {
					rebus[i + 1] = b;
					break;
				}
			} else {
				if (rebus[i] == look_for) {
					rebus[i + 1] = n;
					counter += 1;
				}
			}
		}
	}

	// replacing all other ?'s with 1's
	for (var i = 1; i < rebus.length; i++) {
		if (rebus[i] == '?') {
			rebus[i] = 1;
		}
	}

	rebus = rebus.join(' ');

	return rebus;
};

// Increase 'minus' ?'s
if (count.plus - count.minus > n) {
	var sum = count.plus - n;
	if (sum >= count.minus && sum <= count.minus * n) {
		print('Possible');
		print(getRebus(count.minus, count.plus, minus));
	} else {
		print('Impossible');
	}
// Increase 'plus' ?'s
} else {
	var sum = count.minus + n;
	if (sum >= count.plus && sum <= count.plus * n) {
		print('Possible');
		print(getRebus(count.plus, count.minus, plus));
	} else {
		print('Impossible');
	}
}
