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>
const double pi = 3.141592653589793238462643383279;
using namespace std;
const double EPS = 1E-10;
long long dp1[20][2][4];
long long dp2[20][2][4];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  int q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    l--;
    string s = to_string(r), t = to_string(l);
    memset(dp1, 0, sizeof(dp1));
    memset(dp2, 0, sizeof(dp2));
    dp1[0][1][0] = 1;
    for (int i = 0; i < s.size(); i++) {
      int d = s[i] - '0';
      for (int f = 0; f < 2; f++) {
        if (f == 0) {
          for (int j = 0; j < 10; j++) {
            for (int k = 0; k <= 3; k++) {
              if (j == 0) {
                dp1[i + 1][0][k] += dp1[i][0][k];
              } else {
                if (k == 3) continue;
                dp1[i + 1][0][k + 1] += dp1[i][0][k];
              }
            }
          }
        } else {
          for (int j = 0; j <= d; j++) {
            for (int k = 0; k <= 3; k++) {
              if (j == d) {
                if (j == 0) {
                  dp1[i + 1][1][k] += dp1[i][1][k];
                } else {
                  if (k == 3) continue;
                  dp1[i + 1][1][k + 1] += dp1[i][1][k];
                }
              } else {
                if (j == 0) {
                  dp1[i + 1][0][k] += dp1[i][1][k];
                } else {
                  if (k == 3) continue;
                  dp1[i + 1][0][k + 1] += dp1[i][1][k];
                }
              }
            }
          }
        }
      }
    }
    dp2[0][1][0] = 1;
    for (int i = 0; i < t.size(); i++) {
      int d = t[i] - '0';
      for (int f = 0; f < 2; f++) {
        if (f == 0) {
          for (int j = 0; j < 10; j++) {
            for (int k = 0; k <= 3; k++) {
              if (j == 0) {
                dp2[i + 1][0][k] += dp2[i][0][k];
              } else {
                if (k == 3) continue;
                dp2[i + 1][0][k + 1] += dp2[i][0][k];
              }
            }
          }
        } else {
          for (int j = 0; j <= d; j++) {
            for (int k = 0; k <= 3; k++) {
              if (j == d) {
                if (j == 0) {
                  dp2[i + 1][1][k] += dp2[i][1][k];
                } else {
                  if (k == 3) continue;
                  dp2[i + 1][1][k + 1] += dp2[i][1][k];
                }
              } else {
                if (j == 0) {
                  dp2[i + 1][0][k] += dp2[i][1][k];
                } else {
                  if (k == 3) continue;
                  dp2[i + 1][0][k + 1] += dp2[i][1][k];
                }
              }
            }
          }
        }
      }
    }
    long long a = 0, b = 0;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j <= 3; j++) {
        a += dp1[s.size()][i][j];
        b += dp2[t.size()][i][j];
      }
    }
    cout << a - b << endl;
  }
  return 0;
}