#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5, oo = 0x3f3f3f3f, mod = 1e9 + 7;
int t;
string l, r;
long long mem[19][5][2];
long long dp(int i, int n, int pre, string s, int sz) {
  if (i == sz) {
    if (n <= 3) return 1;
    return 0;
  }
  if (n > 3) return 0;
  if (mem[i][n][pre] != -1) return mem[i][n][pre];
  int lim = 9;
  if (!pre) lim = s[i] - '0';
  long long ans = 0;
  for (int d = 0; d <= lim; d++) {
    int nd = n;
    if (d != 0) nd++;
    int pred = pre;
    if (pre == 0 && d < lim) pred = 1;
    if (nd <= 3) ans += dp(i + 1, nd, pred, s, sz);
  }
  return mem[i][n][pre] = ans;
}
void dec() {
  int i = l.size() - 1;
  while (i >= 0) {
    if (l[i] != '0') {
      l[i]--;
      break;
    } else
      l[i] = '9';
    i--;
  }
}
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    memset(mem, -1, sizeof mem);
    cin >> l >> r;
    dec();
    long long ansR = dp(0, 0, 0, r, r.size());
    memset(mem, -1, sizeof mem);
    cout << ansR - dp(0, 0, 0, l, l.size()) << endl;
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5, oo = 0x3f3f3f3f, mod = 1e9 + 7;
int t;
string l, r;
long long mem[19][5][2];
long long dp(int i, int n, int pre, string s, int sz) {
  if (i == sz) {
    if (n <= 3) return 1;
    return 0;
  }
  if (n > 3) return 0;
  if (mem[i][n][pre] != -1) return mem[i][n][pre];
  int lim = 9;
  if (!pre) lim = s[i] - '0';
  long long ans = 0;
  for (int d = 0; d <= lim; d++) {
    int nd = n;
    if (d != 0) nd++;
    int pred = pre;
    if (pre == 0 && d < lim) pred = 1;
    if (nd <= 3) ans += dp(i + 1, nd, pred, s, sz);
  }
  return mem[i][n][pre] = ans;
}
void dec() {
  int i = l.size() - 1;
  while (i >= 0) {
    if (l[i] != '0') {
      l[i]--;
      break;
    } else
      l[i] = '9';
    i--;
  }
}
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    memset(mem, -1, sizeof mem);
    cin >> l >> r;
    dec();
    long long ansR = dp(0, 0, 0, r, r.size());
    memset(mem, -1, sizeof mem);
    cout << ansR - dp(0, 0, 0, l, l.size()) << endl;
  }
  return 0;
}

```