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.util.Scanner;
  
  public class B {
      static final Reader sc = new Reader();
  
      public static void main(String[] args) throws IOException {
          int t = sc.nextInt();
          while (t-- > 0) {
              //Solve here
              long n = sc.nextLong();
              long k = sc.nextLong();
              System.out.println((int) decToBinary(k, n));
          }
  
      }
  
      static int M = 1_000_000_007;
  
      static long decToBinary(long k, long n) {
          long[] binaryNum = new long[32];
          long M = 1000000007;
          int i = 0;
          while (k > 0) {
              binaryNum[i] = k % 2;
              k = k / 2;
              i++;
          }
          long sum = 0;
          for (int j = 0; j < 32; j++)
              if (binaryNum[j] == 1)
                  sum = (sum + (power(n, j)) % M) % M;
          return sum;
      }
  
      static long power(long x, long y) {
          long res = 1; // Initialize result
          x = x % M; // Update x if it is more than or
          // equal to p
          if (x == 0)
              return 0; // In case x is divisible by p;
          while (y > 0) {
              // If y is odd, multiply x with result
              if ((y & 1) != 0)
                  res = (res * x) % M;
              // y must be even now
              y = y >> 1; // y = y/2
              x = (x * x) % M;
          }
          return res;
      }
  
      static long modMult(long a, long b) {
          return a * b % M;
      }
  
  
      static class Reader {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public Reader() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public Reader(String file_name) throws IOException {
              din = new DataInputStream(new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public String readLine() throws IOException {
              byte[] buf = new byte[64]; // line length
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n')
                      break;
                  buf[cnt++] = (byte) c;
              }
              return new String(buf, 0, cnt);
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ')
                  c = read();
              boolean neg = (c == '-');
              if (neg)
                  c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
  
              if (neg)
                  return -ret;
              return ret;
          }
  
          public long nextLong() throws IOException {
              long ret = 0;
              byte c = read();
              while (c <= ' ')
                  c = read();
              boolean neg = (c == '-');
              if (neg)
                  c = read();
              do {
                  ret = ret * 10 + c - '0';
              }
              while ((c = read()) >= '0' && c <= '9');
              if (neg)
                  return -ret;
              return ret;
          }
  
          public double nextDouble() throws IOException {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ')
                  c = read();
              boolean neg = (c == '-');
              if (neg)
                  c = read();
  
              do {
                  ret = ret * 10 + c - '0';
              }
              while ((c = read()) >= '0' && c <= '9');
  
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
  
              if (neg)
                  return -ret;
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1)
                  buffer[0] = -1;
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead)
                  fillBuffer();
              return buffer[bufferPointer++];
          }
  
          public void close() throws IOException {
              if (din == null)
                  return;
              din.close();
          }
      }
  }
  