#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vpii = vector<pii>;
template <class T>
inline bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, const T& b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const ll infLL = 1e18;
const ld eps = 1e-6;
ll dp[20][10][2][5];
vl nums;
void calc(int p = 0, int z = 0, ll num = 0) {
  if (p == 18) {
    nums.push_back(num);
    return;
  }
  calc(p + 1, z, num * 10);
  if (z < 3)
    for (int i = (1); i < (10); i++) calc(p + 1, z + 1, num * 10 + i);
}
void solve() {
  ll l, r;
  cin >> l >> r;
  cout << upper_bound(nums.begin(), nums.end(), r) -
              lower_bound(nums.begin(), nums.end(), l)
       << nl;
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  calc();
  int t;
  cin >> t;
  nums.push_back(1e18);
  sort(nums.begin(), nums.end());
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in CPP for 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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vpii = vector<pii>;
template <class T>
inline bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, const T& b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const ll infLL = 1e18;
const ld eps = 1e-6;
ll dp[20][10][2][5];
vl nums;
void calc(int p = 0, int z = 0, ll num = 0) {
  if (p == 18) {
    nums.push_back(num);
    return;
  }
  calc(p + 1, z, num * 10);
  if (z < 3)
    for (int i = (1); i < (10); i++) calc(p + 1, z + 1, num * 10 + i);
}
void solve() {
  ll l, r;
  cin >> l >> r;
  cout << upper_bound(nums.begin(), nums.end(), r) -
              lower_bound(nums.begin(), nums.end(), l)
       << nl;
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  calc();
  int t;
  cin >> t;
  nums.push_back(1e18);
  sort(nums.begin(), nums.end());
  while (t--) {
    solve();
  }
}

```