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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.zip.CheckedInputStream;
  
  import javax.naming.InitialContext;
  
  import java.util.*;
  
  public class Main {
  
  	static long mod = 1_000_000_007;
  
  	public static void main(String[] args) {
  
  		FastScanner f = new FastScanner();
  		PrintWriter p = new PrintWriter(System.out);
  		int t=f.nextInt();
  		while(t-->0) {
  			long n=f.nextLong();
  			long k=f.nextLong();
  			long ans=res(n, k);
  			p.println(ans);
  		}
  		p.close();
  
  
  	}
  	
  	static long res(long n,long k) {
  		
  		long curr=1;
  		long ans=0;
  		for(int bit=0;bit<=31;bit++) {
  			if((k&(1<<bit))!=0) {
  				ans=add(ans,curr);
  			}
  			curr=mul(curr, n);
  		}
  		
  		return ans;
  	}
  	
  	
  	
  	
  	
  	
  	
  
  	
  
  //-----------------------------------------------------------------------------------------------------------------------------------------------------------------
  
  	static int offx[] = { -1, -1, -1, 0, 1, 1, 1, 0 };
  	static int offy[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
  
  	static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastScanner() {
  			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 add(long x, long y) {
  		x += y;
  		if (x >= mod)
  			return (x % mod);
  		return x;
  	}
  
  	static long sub(long x, long y) {
  		x -= y;
  		if (x < 0)
  			return (x + mod);
  		return x;
  	}
  
  	static long mul(long x, long y) {
  		return (x * y) % mod;
  	}
  
  	static long bin_pow(long x, long p) {
  		if (p == 0)
  			return 1;
  		if ((p & 1) != 0)
  			return mul(x, bin_pow(x, p - 1));
  		return bin_pow(mul(x, x), p / 2);
  	}
  
  	static long rev(long x) {
  		return bin_pow(x, mod - 2);
  	}
  
  	static long div(long x, long y) {
  		return mul(x, rev(y));
  	}
  
  	// will work till array size of 65537---fastest sorting time
  //	static long[] radixSort(long[] f) {
  //		return radixSort(f, f.length);
  //	}
  //
  //	static long[] radixSort(long[] f, int n) {
  //		long[] to = new long[n];
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 16 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 16 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 32 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 32 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 48 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 48 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		return f;
  //	}
  
  	// Function to sort an array using quick sort algorithm.
  	static void quickSort(long arr[], int low, int high) {
  		if (low < high) {
  			int p = partition(arr, low, high);
  			quickSort(arr, low, p - 1);
  			quickSort(arr, p + 1, high);
  		}
  	}
  
  	static int partition(long arr[], int low, int high) {
  		long pivot = arr[high];
  		int i = low - 1;
  		for (int j = low; j < high; j++) {
  			if (arr[j] < pivot) {
  				i++;
  				long temp = arr[i];
  				arr[i] = arr[j];
  				arr[j] = temp;
  			}
  		}
  		long temp = arr[i + 1];
  		arr[i + 1] = arr[high];
  		arr[high] = temp;
  
  		return i + 1;
  	}
  
  	/// Ceils
  	static long ceil(long x, long m) {
  		long res = x / m;
  		if (x % m != 0) {
  			res++;
  		}
  		return res;
  	}
  
  	// ------------------------------------------------------------------------------------------------
  	// makes the prefix sum array
  	static long[] prefixSum(long arr[], int n) {
  		long psum[] = new long[n];
  		psum[0] = arr[0];
  		for (int i = 1; i < n; i++) {
  			psum[i] = psum[i - 1] + arr[i];
  		}
  		return psum;
  	}
  	// ------------------------------------------------------------------------------------------------
  
  	// makes the suffix sum array
  	static long[] suffixSum(long arr[], int n) {
  		long ssum[] = new long[n];
  		ssum[n - 1] = arr[n - 1];
  		for (int i = n - 2; i >= 0; i--) {
  			ssum[i] = ssum[i + 1] + arr[i];
  		}
  		return ssum;
  	}
  
  //------------------------------------------------------------------------------------------
  	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
  	// OVERHEADS
  //	static long m = (long) (1e9 + 7);
  
  	static long binPower(long a, long n, long m) {
  		if (n == 0)
  			return 1;
  		long res = 1;
  		while (n > 0) {
  			if ((n & 1) != 0) {
  				res *= a;
  			}
  			a *= a;
  			n >>= 1;
  		}
  		return res;
  	}
  
  //-------------------------------------------------------------------------------------------
  
  	// gcd
  	static long gcd(long a, long b) {
  		return b == 0 ? a : gcd(b, a % b);
  	}
  
  //------------------------------------------------------------------------------------------
  
  	// lcm
  	static long lcm(long a, long b) {
  		return a / gcd(a, b) * b;
  	}
  
  	// ------------------------------------------------------------------------------------------
  	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
  	// O(LOGn)
  	static int setBits(int n) {
  		int count = 0;
  		while (n > 0) {
  			n = n & (n - 1);
  			count++;
  		}
  		return count;
  	}
  //------------------------------------------------------------------------------------------
  
  	// ------------------------------------------------------------------------------------------
  	// 0 based indexing
  	static boolean KthBitSet(int n, int k) {
  		int mask = 1;
  		mask = mask <<= k;
  		if ((mask & n) != 0)
  			return true;
  		else
  			return false;
  	}
  
  //------------------------------------------------------------------------------------------
  
  	// EXTENDED EUCLIDEAN THEOREM
  	// TO REPRESENT GCD IN TERMS OF A AND B
  	// gcd(a,b) = a.x + b.y where x and y are integers
  
  	static long x = -1;
  	static long y = -1;
  
  	static long gcdxy(long a, long b) {
  		if (b == 0) {
  			x = 1;
  			y = 0;
  			return a;
  		} else {
  			long d = gcdxy(b, a % b);
  			long x1 = y;
  			long y1 = x - (a / b) * y;
  			x = x1;
  			y = y1;
  
  			return d;
  		}
  	}
  
  //-------------------------------------------------------------------------------------------------F
  }
  