//给定一个数 求他百位上面出现k的次数 自行枚举找规律

//比如求0的次数

//那么分为 =0 和 > 0的情况

//=0  比如 1029 百位 出现0的有 1000-1029 30个 (before-1)*100 + after+1

//>0  比如 1129 百位 出现0的有 1000-1099 100个 (before)*100

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll dp[];

ll Count(ll x,ll k)

{

    ll i =;

    ll res =;

    ll before = , cur= ,after =;

    while ((x/i))

    {

        cur = (x/i)%;

        before = x/(i*);

        after = x-(before*+cur)*i;

        if(cur == k)

            res += before*i +after+;

        else if(cur < k)

            res += before*i;

        else

            res += (before+)*i;

        i*=;

    }

    return res;

}

ll Count0(ll x)

{

    ll i =;

    ll res =;

    ll before = , cur= ,after =;

    while ((x/i))

    {

        cur = (x/i)%;

        before = x/(i*);

        after = x-(before*+cur)*i;

        if(cur == )

            res += (before-)*i +after+;

        else

            res += before*i;

        i*=;

    }

    return res;

}

int main ()

{

    ll a,b;

    cin >> a >> b;

    dp[] = Count0(b) - Count0(a-);

    cout<< dp[] <<endl;

    for(int i=;i<=;i++){//统计出来1 - 9的数量了

        dp[i] = Count(b,i) - Count(a-,i);

        cout<< dp[i] <<endl;

    }

    //差一个数字0的数量

}