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;
long long t, b, l, r;
long long a[60];
long long num[10];
long long dp[11][60][1024];
inline long long get(long long st, long long sh) {
  memset(num, 0, sizeof(num));
  long long cnt = 9, tot = 0;
  while (cnt) {
    num[cnt--] = st % 2;
    st /= 2;
  }
  num[9 - sh]++;
  for (int i = 0; i <= 9; i++) {
    if (num[i] == 2) num[i] = 0;
    tot = tot * 2 + num[i];
  }
  return tot;
}
long long dfs(long long now, long long st, bool limit, bool lead) {
  if (!now) return (!st);
  if (!lead && !limit && dp[b][now][st] != -1) return dp[b][now][st];
  long long up = limit ? a[now] : b - 1;
  long long sum = 0;
  for (int i = 0; i <= up; i++) {
    int new_st = ((i == 0 && lead) ? st : get(st, i));
    sum += dfs(now - 1, new_st, limit && i == a[now], lead && i == 0);
  }
  if (!lead && !limit) dp[b][now][st] = sum;
  return sum;
}
inline long long solve(long long num, long long b) {
  long long pos = 0;
  while (num) {
    a[++pos] = num % b;
    num /= b;
  }
  return dfs(pos, 0, 1, 1);
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld%lld", &b, &l, &r);
    printf("%lld\n", solve(r, b) - solve(l - 1, b));
  }
}
