#include <bits/stdc++.h>
using namespace std;
int t, n;
long long L, R, a, ans;
vector<long long> v;
long long dp[20][4][5];
long long solve(int idx, int cnt, int p) {
  if (cnt > 3) return 0;
  if (idx == n) return 1;
  long long &ret = dp[idx][cnt][p];
  if (ret != -1) return ret;
  long long m = 0;
  if (!p) {
    m = solve(idx + 1, cnt, v[idx] > 0);
    if (v[idx] > 1) m += (v[idx] - 1) * solve(idx + 1, cnt + 1, 1);
    if (v[idx] > 0) m += solve(idx + 1, cnt + 1, 0);
  } else {
    m = solve(idx + 1, cnt, 1);
    m += 9 * solve(idx + 1, cnt + 1, 1);
  }
  return ret = m;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    v.clear();
    a = R;
    while (a > 0) {
      v.push_back(a % 10);
      a /= 10;
    }
    n = v.size();
    reverse(v.begin(), v.end());
    memset(dp, -1, sizeof(dp));
    ans = solve(0, 0, 0);
    v.clear();
    a = L - 1;
    while (a) {
      v.push_back(a % 10);
      a /= 10;
    }
    n = v.size();
    reverse(v.begin(), v.end());
    memset(dp, -1, sizeof(dp));
    ans -= solve(0, 0, 0);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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;
int t, n;
long long L, R, a, ans;
vector<long long> v;
long long dp[20][4][5];
long long solve(int idx, int cnt, int p) {
  if (cnt > 3) return 0;
  if (idx == n) return 1;
  long long &ret = dp[idx][cnt][p];
  if (ret != -1) return ret;
  long long m = 0;
  if (!p) {
    m = solve(idx + 1, cnt, v[idx] > 0);
    if (v[idx] > 1) m += (v[idx] - 1) * solve(idx + 1, cnt + 1, 1);
    if (v[idx] > 0) m += solve(idx + 1, cnt + 1, 0);
  } else {
    m = solve(idx + 1, cnt, 1);
    m += 9 * solve(idx + 1, cnt + 1, 1);
  }
  return ret = m;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d %I64d", &L, &R);
    v.clear();
    a = R;
    while (a > 0) {
      v.push_back(a % 10);
      a /= 10;
    }
    n = v.size();
    reverse(v.begin(), v.end());
    memset(dp, -1, sizeof(dp));
    ans = solve(0, 0, 0);
    v.clear();
    a = L - 1;
    while (a) {
      v.push_back(a % 10);
      a /= 10;
    }
    n = v.size();
    reverse(v.begin(), v.end());
    memset(dp, -1, sizeof(dp));
    ans -= solve(0, 0, 0);
    printf("%I64d\n", ans);
  }
  return 0;
}

```