#include <bits/stdc++.h>
using namespace std;
long long f[21][4], g[21][4], d[21];
long long calc(long long x) {
  long long cnt = 0, re = 0;
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  f[0][0] = g[0][0] = 1;
  while (x) {
    d[++cnt] = x % 10;
    x /= 10;
  }
  for (long long i = 1; i <= cnt; i++)
    for (long long j = 0; j < 4; j++) {
      if (d[i]) {
        if (j > 0) f[i][j] = g[i - 1][j - 1] * (d[i] - 1) + f[i - 1][j - 1];
        f[i][j] += g[i - 1][j];
      } else
        f[i][j] = f[i - 1][j];
      g[i][j] = g[i - 1][j];
      if (j > 0) g[i][j] += 9 * g[i - 1][j - 1];
    }
  for (long long i = 0; i < 4; i++) re += f[cnt][i];
  return re;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
signed main() {
  long long t = read();
  while (t--) {
    long long L = read(), R = read();
    printf("%lld\n", calc(R) - calc(L - 1));
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long f[21][4], g[21][4], d[21];
long long calc(long long x) {
  long long cnt = 0, re = 0;
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  f[0][0] = g[0][0] = 1;
  while (x) {
    d[++cnt] = x % 10;
    x /= 10;
  }
  for (long long i = 1; i <= cnt; i++)
    for (long long j = 0; j < 4; j++) {
      if (d[i]) {
        if (j > 0) f[i][j] = g[i - 1][j - 1] * (d[i] - 1) + f[i - 1][j - 1];
        f[i][j] += g[i - 1][j];
      } else
        f[i][j] = f[i - 1][j];
      g[i][j] = g[i - 1][j];
      if (j > 0) g[i][j] += 9 * g[i - 1][j - 1];
    }
  for (long long i = 0; i < 4; i++) re += f[cnt][i];
  return re;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
signed main() {
  long long t = read();
  while (t--) {
    long long L = read(), R = read();
    printf("%lld\n", calc(R) - calc(L - 1));
  }
  return 0;
}

```