/* 
 * A solution for "The Trip" problem.
 * UVa ID: 10137
 */
#include <stdio.h>

int main (int argc, const char * argv[]) {
	/* number of students in the trip */
	long numOfStudents;
	
	/* the total sum of money spent */
	double total;
	
	/* the total amount of money to exchange in order to equalize */
	double exchange;
	
	/* the equalized trip amount to be payed by each student */
	double equalizedAmount;
	
	/* difference between the equalized amount and the amount spent */
	double diff;
	
	/* sum of all negative differences */
	double negativeSum;
	
	/* sum of all positive differences */
	double positiveSum;
	
	/* iterator */
	int i;
	
	while(scanf("%ld", &numOfStudents) != EOF) {
				
		/* 0, ends the program */
		if (!numOfStudents) {
			return 0;
		}
		
		/* keeps the amount of money spent by each student */
		double amountSpent[numOfStudents];		
		
		/* clean */
		total = 0;
		negativeSum = 0;
		positiveSum = 0;
				
		for (i = 0; i < numOfStudents; i++) {
			scanf("%lf\n", &amountSpent[i]);
			total += amountSpent[i];
		}
				
		equalizedAmount = total / numOfStudents;
		
		for (i = 0; i < numOfStudents; i++) {
			/* to ensure 0.01 precision */
			diff = (double) (long) ((amountSpent[i] - equalizedAmount) * 100.0) / 100.0;
			
			if (diff < 0) {
				negativeSum += diff;
			} else { 
				positiveSum += diff;
			}
		}

		/* when the total amount is even, these sums do not differ. otherwise, they differ in one cent */
		exchange = (-negativeSum > positiveSum) ? -negativeSum : positiveSum;
						
		/* output result */
		printf("$%.2lf\n", exchange);
	}

    return 0;
}

/* 
 * A solution for the Minesweeper problem.
 * UVa ID: 10189
 */
#include <stdio.h>

#define MINE '*'
#define MATRIX_OFFSET 1

int main (int argc, const char * argv[]) {
	/* number of lines of the mine field */
	long n;
	
	/* number of colums of the mine field */
	long m;
	
	/* iterators for retrieving the content of each field line*/
	long i, j;
	
	/* field number identifier */
	long fieldNumber = 0;

	while(scanf("%ld %ld", &n, &m) != EOF) {
			
		/* 0 0, ends the program */
		if (!n && !m) {
			return 0;
		} else {
			if (fieldNumber > 0) {
				/* 
				 * workaround for the extra line on the output, 
				 * which was giving me a Wrong Answer verdict..
				 * so only insert an extra line if there's more
				 * to come :-) 
				 */
				printf("\n");
			}
		}
		
		/* increment field id */
		fieldNumber++;
				
		/* a bidimensional array representing the input mine field */
		char inputField[n + MATRIX_OFFSET + 1][m + MATRIX_OFFSET + 1];
		
		/* a bidimensional array to output the mine field */
		int outputField[n + MATRIX_OFFSET + 1][m + MATRIX_OFFSET + 1];
		
		/* temporary variable to retrieve mine field lines */
		char line[m];
		
		/* init, O(n2) */
		for (i = 0; i < n + MATRIX_OFFSET + 1; i++) {
			for (j = 0; j < m + MATRIX_OFFSET + 1; j++) {
				outputField[i][j] = 0;
			}
		}
		
		/* read mine field, O(n2) */
		for (i = 0; i < n; i++) {
			scanf("%s\n",line);
			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
				inputField[i + MATRIX_OFFSET][j] = line[j - MATRIX_OFFSET];
			}
		}
		
		/* update output, O(n2) */ 
		for (i = MATRIX_OFFSET; i < n + MATRIX_OFFSET; i++) {
			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
				if (inputField[i][j] == MINE) {
				
					/* upper neighbours */
					outputField[i-1][j-1]++;
					outputField[i-1][j]++;
					outputField[i-1][j+1]++;
					
					/* same level neighbours */
					outputField[i][j-1]++;
					outputField[i][j+1]++;
					
					/* lower neighbours */
					outputField[i+1][j-1]++;
					outputField[i+1][j]++;
					outputField[i+1][j+1]++;
				}
			}
		}
		
		/* present output, O(n2) */
		printf("Field #%ld:\n",fieldNumber);
		for (i = MATRIX_OFFSET; i < n + MATRIX_OFFSET; i++) {
			for (j = MATRIX_OFFSET; j < m + MATRIX_OFFSET; j++) {
				if (inputField[i][j] == MINE)
					printf("%c", MINE);
				else	
					printf("%d",outputField[i][j]);
			}
			printf("\n");
		}

	}

    return 0;
}