#include<iostream>
#include<string.h>
#define ll long long
using namespace std;
ll shu[20], dp[20][20][2000];//dp[i][j][k],i是长度，j是支点，k是力矩和,dp[i][j][k]是以j为支点的平衡数的数量
ll dfs(ll len, ll zhidian, ll sum, bool shangxian)
{
    if (len == 0)
        return sum == 0 ? 1 : 0;
    if (sum < 0)//因为力矩和是从支点右边开始算的（sum>0）,如果左边的力矩都处理完之后sum<0,那一定不是平衡数
        return 0;
    if (!shangxian&&dp[len][zhidian][sum]!=-1)
        return dp[len][zhidian][sum];
    ll mx, cnt = 0;
    mx = (shangxian ? shu[len] : 9);
    for (ll i = 0; i <= mx; i++)//注意是<=
    {
        ll temp = sum;
        temp = temp + (len - zhidian)*i;//len会不断的被return ,细细体会
        cnt = cnt + dfs(len - 1, zhidian, temp, i == mx && shangxian);
    }
    if (!shangxian)
        dp[len][zhidian][sum] = cnt;
    return cnt;
}

ll solve(ll n)
{
    ll len = 0;
    while (n)
    {
        shu[++len] = n % 10;
        n = n / 10;
    }
    ll ans = 0;
    for (ll i = 1; i <= len; i++)//支点是从1开始，因为最高位的数一定不是平衡数
        ans = ans + dfs(len, i, 0, true);
    return ans - (len - 1);//如果0是支点，程序也会判断是平衡数，但是不符合题意
}
int main()
{
    ll l, r, t;
    scanf("%lld", &t);
    memset(dp,-1,sizeof(dp));//如果默认0会TLE
    while (t--)
    {
        scanf("%lld%lld", &l, &r);
        printf("%lld\n", solve(r) - solve(l - 1));
    }
    return 0;
}