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.util.*;
  import java.io.*;
  public class T2 {
  
  	static long m = 1000000007;
  	static class fast {
  		BufferedReader br;
  		StringTokenizer st;
  		public fast() {
  			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;
  		}
  	}
  
  	static long binpow( long a,  long b) {
  		long res = 1;
  		while (b > 0) {
  			if ((b & 1) == 1)
  				res = (res * a) % m;
  			a = (a * a) % m;
  			b >>= 1;
  		}
  		return res;
  	}
  
  	public static void main(String[] args)throws IOException {
  		if (System.getProperty("ONLINE_JUDGE") == null) {
  			try {
  				System.setIn(new FileInputStream(new File("i.txt")));
  				System.setOut(new PrintStream(new File("o.txt")));
  			} catch (Exception e) {}
  		}
  		fast f = new fast();
  		int t = f.nextInt();
  		StringBuilder sb = new StringBuilder();
  		while (t-- > 0) {
  			long n = f.nextLong();
  			long k = f.nextLong();
  			long index = 0; long ans = 0;
  			while (k > 0) {
  				if ((k & 1) == 1) {
  					ans += binpow(n, index);
  				}
  				index++;
  				k = (k >> 1);
  			}
  			sb.append((ans) % m + "\n");
  		}
  		System.out.println(sb);
  	}
  }
  
  
  
  
  
  
  
  
  
  
  
  