Just Add It#5493

Just Add It

1 second

128 megabytes

Very easy

For two given integers $n$ and $k$ find

(Z_{n} + Z_{n - 1} - 2 \cdot Z_{n - 2}) m o d 10000007,

(Z_n + Z_{n-1} - 2 \cdot Z_{n-2})~mod~10000007,

where

Z_{n} = S_{n} + P_{n},

Z_n = S_n + P_n,

S_{n} = 1^{k} + 2^{k} + 3^{k} + ... + n^{k},

S_n = 1^k + 2^k + 3^k + ... + n^k,

P_{n} = 1^{1} + 2^{2} + 3^{3} + ... + n^{n}

P_n = 1^1 + 2^2 + 3^3 + ... + n^n

Input

There are several test cases. Each case is a separate line that contains two positive integers $n$ and $k (1 < n < 2 \cdot 1 0^{8}, 0 < k < 1 0^{6})$ k~(1 < n < 2 \cdot 10^8, 0 < k < 10^6). The last line contains two zeros and should not be processed.

Output

For each test case print the required value in a separate line.

Examples

Example 1

Loading

(Z_{n} + Z_{n - 1} - 2 \cdot Z_{n - 2}) m o d 10000007,

Z_{n} = S_{n} + P_{n},

S_{n} = 1^{k} + 2^{k} + 3^{k} + ... + n^{k},

P_{n} = 1^{1} + 2^{2} + 3^{3} + ... + n^{n}

k (1 < n < 2 \cdot 1 0^{8}, 0 < k < 1 0^{6})