# 消成上三角矩阵
# 1.枚举列
# 2.找非零行
# 3.交换
# 4.下面消零

# 判断解的三种情况

def gauss(n):
    r = 0
    for c in range(n):
        t = r
        #找到非0行
        for i in range(r,n):
            #如果i行c列不是0
            if a[i][c] :
                t = i
                break

        #如果找不到i行c列不是0
        if not a[t][c]: continue

        #如果找到非零行，那么和当前行交换
        for i in range(c,n+1):a[t][i],a[r][i] = a[r][i],a[t][i]

        #然后用r行把下面所有行的当前列消成0
        for i in range(r+1,n):
            #如果i行c列不是0
            if a[i][c]:
                #那么从第c列开始消
                for j in range(c,n+1):
                    a[i][j] ^= a[r][j]

        r += 1

    if r < n:
        for i in range(r,n):
            #如果等号右边不是零 那么无解
            if a[i][n]:
                return 2
        #如果后面行全是零，那么无穷解
        return 1

    #如果有解那么从下往上消元
    for i in range(n,-1,-1):
        for j in range(i+1,n):
            #a[j][n]是j行的解
            a[i][n] ^= a[i][j] & a[j][n]

    return 0

if __name__ == "__main__":

    n = int(input())
    a = [0]*n
    for i in range(n):
        a[i] = list(map(int,input().split()))

    res = gauss(n)

    if res == 1:
        print('Multiple sets of solutions')
    elif res == 2:
        print('No solution')
    else:
        for i in range(n): print(a[i][n])