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#include <cstdio>
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#include <cstring>
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#include <iostream>
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#include <cmath>
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using namespace std;
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typedef long long LL;
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const int N = 20;
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/*
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x,y <= 10^{18},用10进制拆分到数组a,20位足够
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本题对于内存空间的限制比较极限，如果无脑使用了32，就会使得f的容量超限
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32*32*10010=10250240 byte = 10010 kb = 9.775390625 mb
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内存上限是：65535kb=64mb
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因为C++本身运行还需要一次的内存，64MB是很极限的，所以在多维数组创建时，一定要注意维度的上界
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*/
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LL a[N];
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/*
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pos:走到哪个数位，长度最长是N=20
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k:以哪个数位为平衡数位，可以是1到20
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sum:前面数位的数乘以偏移距离的和,最大偏移量是20，数位最大是9
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可以视为
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10*20+10*19+....+10*1=10*(1+2+3+...+20)=10*21*10=2100
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*/
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LL f[N][N][2110];
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LL dfs(int pos, int k, int sum, bool lead, bool limit) {
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if (pos == 0) return sum == 0;
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if (sum < 0) return 0;
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if (!lead && !limit && ~f[pos][k][sum]) return f[pos][k][sum];
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int up = limit ? a[pos] : 9;
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LL ans = 0;
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for (int i = 0; i <= up; i++)
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// sum + (pos - k) * i : 这句话很牛X，pos > k 是加，pos < k是减
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ans += dfs(pos - 1, k, sum + (pos - k) * i, lead && i == 0, limit && i == a[pos]);
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if (!limit && !lead) return f[pos][k][sum] = ans;
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return ans;
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}
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LL calc(LL x) {
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if (x == -1) return 0; //如果输入的是l=0,则返回0
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int al = 0;
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while (x) a[++al] = x % 10, x /= 10;
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LL ans = 0;
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for (int i = al; i; i--) //枚举每一个数位，模拟此位置k为平衡数位
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ans += dfs(al, i, 0, true, true);
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// 00,000,0000......
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return ans - al + 1;
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}
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(0);
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//平衡数是数的本身性质，与给定的区间无关,可以将memset放在循环外初始化
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memset(f, -1, sizeof f);
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int T;
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cin >> T;
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LL l, r;
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while (T--) {
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cin >> l >> r;
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printf("%lld\n", calc(r) - calc(l - 1));
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}
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return 0;
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}
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