#include <bits/stdc++.h>
using namespace std;
const int N = 60;
int w[N];
int he[N], ver[N], ne[N];
long long f[N][4], val[N];
int su[N];
int tot;
long long ans;
long long dp(int cur, int pre, int flag) {
  if (cur < 0 && pre >= 0) return 1;
  if (cur < 0 || pre < 0) return 0;
  if (!flag && f[cur][pre] != -1) return f[cur][pre];
  int gg = flag ? su[cur] : 9;
  long long ans = 0;
  for (int i = 0; i <= gg; i++) {
    if (i == 0)
      ans += dp(cur - 1, pre, flag && i == gg);
    else
      ans += dp(cur - 1, pre - 1, flag && i == gg);
  }
  if (!flag) f[cur][pre] = ans;
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(f, -1, sizeof(f));
    long long a, b;
    scanf("%lld%lld", &a, &b);
    int len = 0;
    while (b) {
      su[len++] = b % 10;
      b /= 10;
    }
    long long gg = dp(len - 1, 3, 1);
    len = 0;
    a--;
    while (a) {
      su[len++] = a % 10;
      a /= 10;
    }
    long long kk = dp(len - 1, 3, 1);
    printf("%lld\n", gg - kk);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 60;
int w[N];
int he[N], ver[N], ne[N];
long long f[N][4], val[N];
int su[N];
int tot;
long long ans;
long long dp(int cur, int pre, int flag) {
  if (cur < 0 && pre >= 0) return 1;
  if (cur < 0 || pre < 0) return 0;
  if (!flag && f[cur][pre] != -1) return f[cur][pre];
  int gg = flag ? su[cur] : 9;
  long long ans = 0;
  for (int i = 0; i <= gg; i++) {
    if (i == 0)
      ans += dp(cur - 1, pre, flag && i == gg);
    else
      ans += dp(cur - 1, pre - 1, flag && i == gg);
  }
  if (!flag) f[cur][pre] = ans;
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(f, -1, sizeof(f));
    long long a, b;
    scanf("%lld%lld", &a, &b);
    int len = 0;
    while (b) {
      su[len++] = b % 10;
      b /= 10;
    }
    long long gg = dp(len - 1, 3, 1);
    len = 0;
    a--;
    while (a) {
      su[len++] = a % 10;
      a /= 10;
    }
    long long kk = dp(len - 1, 3, 1);
    printf("%lld\n", gg - kk);
  }
  return 0;
}

```