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: import java.io.*;
  import java.lang.reflect.Array;
  import java.math.BigInteger;
  import java.util.*;
  
  public class TT {
  
      static Uzi uzi;
      static int counter = 0;
      static long[] dp;
      static ArrayList<Integer> list;
      static long N = 1000000000 + 7;
      public static void main(String[] args) throws IOException {
          FastReader fk = new FastReader();
  //        Scanner fk = new Scanner(System.in);
          PrintWriter pr = new PrintWriter(System.out);
  //        FileWriter wr = new FileWriter("large_fib_output.txt");
  //        uzi = new Uzi();
  
          int t = fk.nextInt();
  
          while(t-- > 0) {
              long n = fk.nextLong(), k = fk.nextLong();
              long term = fxn(k ,n);
              pr.println(term);
  
          }
  
          pr.close();
  
      }
  
      private static long fxn(long k, long n){
          long sum =0;
          long x = 1;
          while(k != 0){
              if((k & 1) == 1){
                  sum += x;
                  sum = sum % N;
              }
              x = (n * x) % N;
              k >>= 1;
          }
  
          return sum % N;
      }
  
  
  
  }
  
  class Uzi {
      boolean[] primes;
      int N;
      long[] fact;
  
      public Uzi() {
      }
  
      public Uzi(int n) {
          this.N = n;
          this.primes = new boolean[N + 1];
          this.fact = new long[N + 1];
      }
  
      public int fast_exponentiation(int n, int b, int mod) {
          // Time complexity: O(logn), Space complexity: O(1)
          int res = 1;
          while (b > 0) {
              res = (res * (b % 2 == 1 ? n : 1)) % mod;
              n = (n * n) % mod;
              b >>= 1;
          }
          return res;
  
      }
  
      public void fact() {
          fact[0] = 1;
  
          for (int i = 1; i <= N; i++) {
              fact[i] = i * fact[i - 1];
          }
      }
  
      public long gcd(long a, long b) {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
  
      public void seive() {
          Arrays.fill(primes, true);
          primes[0] = primes[1] = false;
  
          for (int i = 2; i * i <= N; i++) {
              if (primes[i]) {
                  for (int j = i * i; j <= N; j += i)
                      primes[j] = false;
              }
          }
      }
  
      public int countDivisors(long num) {
          int counter = 0;
  
          for (int i = 1; i*i <= num ; i++) {
              if (num % i == 0)
                  counter += (i * i == num) ? 1 : 2;
          }
          return counter;
      }
  
      public boolean binary_search(int[] arr, long key) {
          int s = 0, e = arr.length - 1, curr;
  
          while (s <= e) {
              curr = (s + e) / 2;
              if (arr[curr] == key)
                  return true;
              else {
                  if (arr[curr] > key)
                      e = curr - 1;
                  else
                      s = curr + 1;
              }
          }
  
          return false;
      }
  
      private int elementJustGreater(int[] arr, int k) {
          // to find the first arr[i] >= k using binary search
  
          int s = 0, e = arr.length - 1, curr;
          int ans = -1;
  
          while (s <= e) {
              curr = (s + e) / 2;
              if (arr[curr] >= k) {
                  ans = arr[curr];
                  e = curr - 1;
              } else
                  s = curr + 1;
          }
  
          return ans;
      }
  
      public int[] elementJustSmaller(int[] arr, long k) {
          // to find the first arr[i] <= k using binary search
          int s = 0, e = arr.length - 1, curr;
          int[] ans = {-1, 0};
  
          while (s <= e) {
              curr = (s + e) / 2;
              if (arr[curr] <= k) {               // should be less than or equal to
                  //ans = arr[curr];
                  if (arr[curr] == k)
                      ans = new int[]{curr, 0};
                  else
                      ans = new int[]{curr, 1};
                  s = curr + 1;
              } else
                  e = curr - 1;
          }
          return ans;
      }
  }
  
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
  
      public FastReader() {
          br = new BufferedReader(
                  new InputStreamReader(System.in));
      }
  
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int nextInt() {
          return Integer.parseInt(next());
      }
  
      long nextLong() {
          return Long.parseLong(next());
      }
  
      double nextDouble() {
          return Double.parseDouble(next());
      }
  
      String nextLine() {
          String str = "";
          try {
              str = br.readLine();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
  }
  
  
  
  