#include <bits/stdc++.h>
using namespace std;
const long double E = 2.71828182845904523536;
const long double pi = acos(-1);
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const long long inf = 1LL << 30;
const int N = 24;
long long x, mem[N][2][4];
int a[N];
void init(long long x) {
  memset(a, 0, sizeof a);
  int i = 1;
  while (x) {
    a[i++] = x % 10LL;
    x /= 10LL;
  }
}
long long dp(int i, bool g, int cnt) {
  if (cnt > 3) return 0;
  if (i == 0) return 1;
  long long &ret = mem[i][g][cnt];
  if (ret != -1) return ret;
  ret = 0;
  for (int j = 0; j <= (g ? 9 : a[i]); j++)
    ret += dp(i - 1, g | (j < a[i]), cnt + (j != 0));
  return ret;
}
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%lld", &x);
    x--;
    init(x);
    memset(mem, -1, sizeof mem);
    long long L = dp(N - 2, 0, 0);
    scanf("%lld", &x);
    init(x);
    memset(mem, -1, sizeof mem);
    long long R = dp(N - 2, 0, 0);
    printf("%lld\n", R - L);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 long double E = 2.71828182845904523536;
const long double pi = acos(-1);
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const long long inf = 1LL << 30;
const int N = 24;
long long x, mem[N][2][4];
int a[N];
void init(long long x) {
  memset(a, 0, sizeof a);
  int i = 1;
  while (x) {
    a[i++] = x % 10LL;
    x /= 10LL;
  }
}
long long dp(int i, bool g, int cnt) {
  if (cnt > 3) return 0;
  if (i == 0) return 1;
  long long &ret = mem[i][g][cnt];
  if (ret != -1) return ret;
  ret = 0;
  for (int j = 0; j <= (g ? 9 : a[i]); j++)
    ret += dp(i - 1, g | (j < a[i]), cnt + (j != 0));
  return ret;
}
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%lld", &x);
    x--;
    init(x);
    memset(mem, -1, sizeof mem);
    long long L = dp(N - 2, 0, 0);
    scanf("%lld", &x);
    init(x);
    memset(mem, -1, sizeof mem);
    long long R = dp(N - 2, 0, 0);
    printf("%lld\n", R - L);
  }
  return 0;
}

```