#include<iostream>

 #include<algorithm>

 #include<cstring>

 using namespace std;

 int l, r;

 int a[], b[];

 void solve(int n, int m) {

     int x = n / ;//除了末位以外的                     x=297

     int y = n % ;//最后一位的数字范围                  y=4

     int tmp = x;

     //计算小于10*x+1到10*x+y中个位数的个数,b[1-y]也就是（2971-2974）个位数

     for (int i = ; i <= y; i++)b[i] += m;

     //计算10*x-10到10*x-1中个位数的总个数,296 0-296 9 个位数个数每个都是296个

     for (int i = ; i <= ; i++)b[i] += m*x;

     while (tmp) {

     //10*x 到10*x+y非个位的个数,因为从0开始，所以*（y+1）就是非个位位数，当然还要乘m

         b[tmp % ] += m*(y + );

         tmp /= ;

     }

     if (x) solve(x - , m * );

 }

 int main() {

     while (cin >> l >> r && (l || r)) {

         memset(b, , sizeof(b));

         if (l > r)swap(l, r);

         solve(l - , );

         for (int i = ; i <= ; i++)a[i] = b[i], b[i] = ;

         solve(r, );

         cout << b[] - a[];

         for (int i = ; i <= ; i++)cout << " " << b[i] - a[i];

         cout << endl;

     }

     return ;

 }