Problem:
Score: 100 points

Problem Statement

We have two distinct integers A and B.

Print the integer K such that |A - K| = |B - K|.

If such an integer does not exist, print IMPOSSIBLE instead.

Constraints

All values in input are integers.

0 \leq A,\ B \leq 10^9

A and B are distinct.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the integer K satisfying the condition.

If such an integer does not exist, print IMPOSSIBLE instead.

Sample Input 1

2 16

Sample Output 1

9

|2 - 9| = 7 and |16 - 9| = 7, so 9 satisfies the condition.

Sample Input 2

0 3

Sample Output 2

IMPOSSIBLE

No integer satisfies the condition.

Sample Input 3

998244353 99824435

Sample Output 3

549034394

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

long int min(long int a, long int b) {
	if (a < b) {
		return a;
	}
	else {
		return b;
	}
}

int main(void) {
	long int n, i, ans;
	long int attack1;
	long int attack2;

	scanf("%d", &n);
	long int a[1000000], b[1000000] = { 0 };
	for (i = 1; i <= n + 1; i++) {
		scanf("%ld", &a[i]);
	}
	for (i = 1; i <= n + 1; i++) {
		scanf("%ld", &b[i]);
	}

	ans = 0;

	//minは小さいほうを返す。同値の時は、bを返す操作

	attack1 = min(a[1], b[1]);
	attack2 = min(a[2], b[1] - attack1);
	ans = ans + (attack1 + attack2);

	for (i = 2; i <= n; i++) {
		attack1 = min(a[i] - attack2, b[i]);
		attack2 = min(a[i + 1], b[i] - attack1);

		ans += (attack1 + attack2);
	}

	printf("%ld\n", ans);
	return 0;
}