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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dp[25][5][2], num[25];
int dfs(int x, int sum, int limit) {
  if (x == 0) return 1;
  if (!limit && dp[x][sum][limit] != -1) return dp[x][sum][limit];
  if (sum >= 3) return 1;
  int up = limit ? num[x] : 9;
  int tmp = 0;
  for (int i = 1; i <= up; i++) {
    tmp += dfs(x - 1, sum + 1, limit & (i == num[x]));
  }
  tmp += dfs(x - 1, sum, limit & (0 == num[x]));
  if (!limit) dp[x][sum][limit] = tmp;
  return tmp;
}
int ask(long long x) {
  int pos = 0;
  if (!x) return 1;
  while (x) {
    num[++pos] = x % 10;
    x /= 10;
  }
  return dfs(pos, 0, 1);
}
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 0; i <= 20; i++) {
    for (int j = 0; j <= 4; j++) {
      dp[i][j][0] = -1;
      dp[i][j][1] = -1;
    }
  }
  while (T--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%d\n", ask(r) - ask(l - 1));
  }
  return 0;
}