Problem: Let's call some positive integer classy if its decimal representation contains no more than 3 non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000 are not.

You are given a segment [L; R]. Count the number of classy integers x such that L ≤ x ≤ R.

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^4) — the number of segments in a testcase.

Each of the next T lines contains two integers L_i and R_i (1 ≤ L_i ≤ R_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of classy integers on a segment [L_i; R_i].

Example

Input

4
1 1000
1024 1024
65536 65536
999999 1000001


Output

1000
1
0
2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int LINF = 0x3f3f3f3f3f3f3f3f;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int DP[20][5][2];
int num[22];
int call(int pos, int cnt, int flag) {
  if (cnt == 4) return 0;
  if (pos == 0) return 1;
  int &ans = DP[pos][cnt][flag];
  if (ans != -1) return ans;
  ans = 0;
  if (flag) {
    for (int i = 0; i <= 9; i++) {
      int new_cnt = cnt + (i > 0 ? 1 : 0);
      ans += call(pos - 1, new_cnt, flag);
    }
  } else {
    for (int i = 0; i <= num[pos]; i++) {
      int new_cnt = cnt + (i > 0 ? 1 : 0);
      ans += call(pos - 1, new_cnt, num[pos] > i ? 1 : 0);
    }
  }
  return ans;
}
int cal(long long k) {
  int siz = 0;
  while (k) {
    num[++siz] = k % 10;
    k /= 10;
  }
  memset(DP, -1, sizeof(DP));
  ;
  return call(siz, 0, 0);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  while (n--) {
    long long x, y;
    cin >> x >> y;
    int ans = cal(y) - cal(x - 1);
    cout << ans << '\n';
  }
}