10
4 108 114
5 30 155
8 193 197
9 71 169
2 163 166
8 120 144
8 22 151
4 21 166
2 46 127
8 38 51
 20
9 142 172
7 132 256
8 245 315
9 496 496
3 345 362
8 13 162
5 342 470
9 16 488
9 467 482
4 471 478
10 92 224
6 228 261
9 54 167
2 402 409
5 9 10
7 225 360
3 438 483
6 252 342
3 491 493
6 296 367
 1
8 1234567890123 123456789012345

0
3
0
1
3
0
5
16
9
1
 0
0
0
0
4
6
13
7
0
0
1
2
3
0
0
2
12
9
1
4
 774752389568

#include <bits/stdc++.h>
using namespace std;
const int N = 70;
int T, b, cnt;
int a[N];
long long L, R;
long long dp[N][11][(1 << 11) - 1][2][2];
long long dfs(int cur, int sta, bool f, bool g) {
  if (cur == -1) return (sta == (1 << b) - 1) ? 1 : 0;
  if (dp[cur][b][sta][f][g] != -1) return dp[cur][b][sta][f][g];
  int v = b - 1;
  if (f) v = a[cur];
  long long ans = 0;
  for (int i = 0; i <= v; i++) {
    if (g) {
      if (i == 0)
        ans += dfs(cur - 1, sta, f && (i == v), 1);
      else
        ans += dfs(cur - 1, sta ^ (1 << i), f && (i == v), 0);
    } else
      ans += dfs(cur - 1, sta ^ (1 << i), f && (i == v), 0);
  }
  return dp[cur][b][sta][f][g] = ans;
}
long long solve(long long x) {
  cnt = 0;
  while (x) {
    a[cnt++] = x % b;
    x /= b;
  }
  return dfs(cnt - 1, (1 << b) - 1, 1, 1);
}
int main() {
  cin >> T;
  memset(dp, -1, sizeof dp);
  while (T--) {
    scanf("%d%lld%lld", &b, &L, &R);
    printf("%lld\n", solve(R) - solve(L - 1));
  }
}
