#include <bits/stdc++.h>
using namespace std;
string l, r;
int n;
long int dp[21][21][2];
int nonZero(int num) {
  if (num == 0) return 0;
  return 1;
}
long int solve(int idx, int nonZeroNumber, int tight, string l) {
  if (idx == n) {
    if (nonZeroNumber <= 3)
      return 1;
    else
      return 0;
  }
  if (nonZeroNumber > 3) return 0;
  long int ans = 0, &ret = dp[idx][nonZeroNumber][tight];
  if (ret != -1) return ret;
  if (tight) {
    for (int i = 0; i <= l[idx] - '0'; i++) {
      if (i != l[idx] - '0') {
        int type = nonZero(i);
        ans += solve(idx + 1, nonZeroNumber + type, 0, l);
      } else {
        int type = nonZero(i);
        ans += solve(idx + 1, nonZeroNumber + type, 1, l);
      }
    }
  } else {
    for (int i = 0; i <= 9; i++) {
      int type = nonZero(i);
      ans += solve(idx + 1, nonZeroNumber + type, 0, l);
    }
  }
  ret = ans;
  return ret;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    cin >> l >> r;
    long int cnt1, cnt2;
    n = r.length();
    memset(dp, -1, sizeof(dp));
    cnt2 = solve(0, 0, 1, r);
    n = l.length();
    memset(dp, -1, sizeof(dp));
    cnt1 = solve(0, 0, 1, l);
    int ct = 0;
    for (int i = 0; i < n; i++) {
      if (l[i] - '0' > 0) ct++;
    }
    if (ct <= 3) {
      cnt1--;
    }
    printf("%ld\n", cnt2 - cnt1);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
string l, r;
int n;
long int dp[21][21][2];
int nonZero(int num) {
  if (num == 0) return 0;
  return 1;
}
long int solve(int idx, int nonZeroNumber, int tight, string l) {
  if (idx == n) {
    if (nonZeroNumber <= 3)
      return 1;
    else
      return 0;
  }
  if (nonZeroNumber > 3) return 0;
  long int ans = 0, &ret = dp[idx][nonZeroNumber][tight];
  if (ret != -1) return ret;
  if (tight) {
    for (int i = 0; i <= l[idx] - '0'; i++) {
      if (i != l[idx] - '0') {
        int type = nonZero(i);
        ans += solve(idx + 1, nonZeroNumber + type, 0, l);
      } else {
        int type = nonZero(i);
        ans += solve(idx + 1, nonZeroNumber + type, 1, l);
      }
    }
  } else {
    for (int i = 0; i <= 9; i++) {
      int type = nonZero(i);
      ans += solve(idx + 1, nonZeroNumber + type, 0, l);
    }
  }
  ret = ans;
  return ret;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    cin >> l >> r;
    long int cnt1, cnt2;
    n = r.length();
    memset(dp, -1, sizeof(dp));
    cnt2 = solve(0, 0, 1, r);
    n = l.length();
    memset(dp, -1, sizeof(dp));
    cnt1 = solve(0, 0, 1, l);
    int ct = 0;
    for (int i = 0; i < n; i++) {
      if (l[i] - '0' > 0) ct++;
    }
    if (ct <= 3) {
      cnt1--;
    }
    printf("%ld\n", cnt2 - cnt1);
  }
  return 0;
}

```