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;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
constexpr ld EPS = 1e-12;
constexpr int INF = numeric_limits<int>::max() / 2;
constexpr int MOD = 1e9 + 7;
template <typename T>
void printv(const vector<T> &v) {
  int sz = v.size();
  for (int i = 0; i < sz; i++) {
    cout << v[i] << " \n"[i == sz - 1];
  }
}
ll calc(ll x) {
  string s = to_string(x);
  int len = s.size();
  vector<vector<vector<ll>>> dp(len + 1,
                                vector<vector<ll>>(5, vector<ll>(2, 0)));
  dp[0][0][1] = 1;
  for (int i = 0; i < len; i++) {
    for (int j = 0; j <= 3; j++) {
      if (s[i] == '0') {
        dp[i + 1][j][0] += dp[i][j][0];
        dp[i + 1][j + 1][0] += dp[i][j][0] * 9;
        dp[i + 1][j][1] += dp[i][j][1];
      } else {
        int num = s[i] - '0';
        dp[i + 1][j][0] += dp[i][j][0] + dp[i][j][1];
        dp[i + 1][j + 1][0] += dp[i][j][0] * 9 + dp[i][j][1] * (num - 1);
        dp[i + 1][j + 1][1] += dp[i][j][1];
      }
    }
  }
  ll ret = 0;
  for (int j = 0; j <= 3; j++) {
    ret += dp[len][j][0] + dp[len][j][1];
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    ll l, r;
    cin >> l >> r;
    cout << calc(r) - calc(l - 1) << endl;
  }
}