human
PYTHON3 solution for "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...]":
assistant
#Here is Python code to find the k-th special number if they are sorted in increasing order.
 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))