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: for _ in range(int(input())):
      n,k=map(int,input().split())
      m=1e9+7;i=0;ans=0
      MOD = 1000000007;
  
  
      # Returns modulo exponentiation
      # for two numbers represented as
      # long long int. It is used by
      # powerStrings(). Its complexity
      # is log(n)
      def powerLL(x, n):
  
          result = 1;
          while (n):
              if (n & 1):
                  result = result * x % MOD;
              n = int(n / 2);
              x = x * x % MOD;
          return result;
      def powerStrings(sa, sb):
  
          a = 0;
          b = 0;
  
          for i in range(len(sa)):
              a = (a * 10 + (ord(sa[i]) -
                             ord('0'))) % MOD;
  
          for i in range(len(sb)):
              b = (b * 10 + (ord(sb[i]) -
                             ord('0'))) % (MOD - 1);
  
          # exponentiation
          return powerLL(a, b);
      while k>0:
          if k&1:
              ans=(ans+powerStrings(str(n),str(i)))%m
          k=k>>1
          i+=1
      print(int(ans))
  
  