#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 5010;
char str1[N], str2[N];
int p[N], a[N], mod2;
ll dp[N][65][65];
ll dfs(int pos, int sum, int last, bool limit)
{
if (pos < 1)
return !last;
if (!limit && dp[pos][sum][last] != -1)
return dp[pos][sum][last];
ll ans = 0;
int up = (limit ? a[pos] : 9);
for (int i = 0; i <= up; i++)
ans += dfs(pos - 1, (sum + i) % mod2, ((last + i * sum - i * p[pos - 1]) % mod2 + mod2) % mod2,i == up && limit);
ans %= mod;
if (!limit)
dp[pos][sum][last] = ans;
return ans;
}
ll solve(char s[])
{
int len = strlen(s + 1);
for (int i = 1; i <= len; i++)
for (int j = 0; j <= mod2; j++)
for (int k = 0; k <= mod2; k++)
dp[i][j][k] = -1;
for (int i = 1; i <= len; i++)
a[i] = s[len - i + 1] - '0';
return dfs(len, 0, 0, true);
}
int main()
{
p[0] = 1;
int T;
scanf("%d",&T);
while (T--)
{
scanf("%s",str1 + 1);
getchar();
scanf("%s",str2 + 1);
getchar();
scanf("%d",&mod2);
int n = strlen(str2 + 1);
int m = strlen(str1 + 1);
for (int i = 1; i <= n; i++)
p[i] = p[i - 1] * 10 % mod2;
//处理好前缀和的形式,不然下一步忒麻烦了。
str1[m]--;
for (int i = m; i >= 1; i--)
{
//特判，借位
if (str1[i] - '0' < 0)
str1[i] += 10, str1[i - 1]--;
else
break;
}
ll ans= (solve(str2) - solve(str1) + mod) % mod;
printf("%lld
",ans);
}
return 0;
}