#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
// constexpr int MOD = 998244353;
constexpr int DY4[]{1, 0, -1, 0}, DX4[]{0, -1, 0, 1};
constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1};
constexpr int DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T& a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U>
inline bool chmin(T& a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;

long long floor_sum(long long a, long long b, long long m, long long n) {
  long long res = 0;
  if (a < 0) {
    long long nxt_a = a % m;
    if (nxt_a < 0) nxt_a += m;
    res -= n * (n - 1) / 2 * ((nxt_a - a) / m);
    a = nxt_a;
  }
  if (b < 0) {
    long long nxt_b = b % m;
    if (nxt_b < 0) nxt_b += m;
    res -= n * ((nxt_b - b) / m);
    b = nxt_b;
  }
  while (true) {
    if (a >= m) {
      res += n * (n - 1) / 2 * (a / m);
      a %= m;
    }
    if (b >= m) {
      res += n * (b / m);
      b %= m;
    }
    const long long y_max = a * n + b;
    if (y_max < m) break;
    b = y_max % m;
    n = y_max / m;
    std::swap(a, m);
  }
  return res;
}

ll p10(int ex) {
  ll res = 1;
  for (; ex > 0; --ex) res *= 10;
  return res;
}

// |{i \in {1, 2, ..., N} | x <= iK mod D}|
int g(const int n, const ll x, const int k, const ll d) {
  return n - (floor_sum(k, k, d, n) - floor_sum(k, k - x, d, n));
}

constexpr int D = 9;
int f(const int r, const int k, const int c) {
  const int n = r / k;
  if (n <= 0) return 0;
  int ans = 0;
  REP(d, D) {
    // c * 10^d <= ik mod 10^(d + 1) < (c + 1) * 10^d
    ans += g(n, c * p10(d), k, p10(d + 1)) - g(n, (c + 1) * p10(d), k, p10(d + 1));
  }
  return ans;
}

void solve() {
  int l, r, k, c; cin >> l >> r >> k >> c;
  cout << f(r, k, c) - f(l - 1, k, c) << '\n';
}

int main() {
  int t; cin >> t;
  while (t--) solve();
  return 0;
}