description: Theofanis really likes sequences of positive integers, thus his teacher (Yeltsa Kcir) gave him a problem about a sequence that consists of only special numbers.
  
  Let's call a positive number special if it can be written as a sum of different non-negative powers of n. For example, for n = 4 number 17 is special, because it can be written as 4^0 + 4^2 = 1 + 16 = 17, but 9 is not.
  
  Theofanis asks you to help him find the k-th special number if they are sorted in increasing order. Since this number may be too large, output it modulo 10^9+7.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first and only line of each test case contains two integers n and k (2 ≤ n ≤ 10^9; 1 ≤ k ≤ 10^9).
  
  Output
  
  For each test case, print one integer — the k-th special number in increasing order modulo 10^9+7.
  
  Example
  
  Input
  
  
  3
  3 4
  2 12
  105 564
  
  
  Output
  
  
  9
  12
  3595374
  
  Note
  
  For n = 3 the sequence is [1,3,4,9...]
solution: #include <bits/stdc++.h>
  using namespace std;
  const long double EPS = 1e-9;
  const long long INF = 1e18;
  const long long mod = 1e9 + 7;
  bool prime[100002];
  long long fr[200005];
  vector<int> adj[200005];
  int euler[400020];
  void sieve() {
    memset(prime, true, sizeof(prime));
    prime[1] = false;
    for (int i = 2; i <= 100000; i++) {
      if (!prime[i]) continue;
      for (int j = 2 * i; j <= 100000; j += i) {
        prime[j] = false;
      }
    }
  }
  long long factorial(long long n) {
    if (n == 0) return 1;
    long long x = 1;
    for (long long i = 1; i <= n; i++) {
      x = (x * i) % mod;
    }
    return x;
  }
  long long power(long long x, long long y) {
    long long res = 1;
    x = x % mod;
    while (y > 0) {
      if (y & 1) res = (res * x) % mod;
      x = (x * x) % mod;
      y = y >> 1;
    }
    return res % mod;
  }
  long long modInverse(long long n) { return power(n, mod - 2); }
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return ((fr[n] % mod) *
            (modInverse(((fr[r] % mod) * (fr[n - r] % mod)) % mod) % mod) % mod);
  }
  void f() {
    fr[0] = 1;
    fr[1] = 1;
    for (long long i = 2; i <= 200001; i++) {
      fr[i] = (fr[i - 1] * i) % mod;
    }
  }
  void EULER(int k, int &ind, int par) {
    euler[ind++] = k;
    for (auto &x : adj[k]) {
      if (x != par) {
        EULER(x, ind, k);
        euler[ind++] = k;
      }
    }
  }
  long long power1(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      x = (x * x);
      y = y >> 1;
    }
    return res;
  }
  void test_case() {
    long long n, k;
    cin >> n >> k;
    if (n > 2) {
      long long ans = 0;
      for (int i = 0; i <= 30; i++) {
        if ((1 << i) & k) ans = (ans + power(n, (long long)i)) % mod;
      }
      cout << ans << "\n";
    } else {
      cout << k << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test_cases = 1, tt = 1;
    cin >> test_cases;
    while (test_cases--) {
      test_case();
    }
    return 0;
  }
  