Ralph is in the Binary Country. The Binary Country consists of *n* cities and (*n* - 1) bidirectional roads connecting the cities. The roads are numbered from 1 to (*n* - 1), the *i*-th road connects the city labeled <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16be0dab327b8800b1f1feefd23d6dd915b97ac4.png" style="max-width: 100.0%;max-height: 100.0%;"/> (here ⌊ *x*⌋ denotes the *x* rounded down to the nearest integer) and the city labeled (*i* + 1), and the length of the *i*-th road is *L*<sub class="lower-index">*i*</sub>.

Now Ralph gives you *m* queries. In each query he tells you some city *A*<sub class="lower-index">*i*</sub> and an integer *H*<sub class="lower-index">*i*</sub>. He wants to make some tours starting from this city. He can choose any city in the Binary Country (including *A*<sub class="lower-index">*i*</sub>) as the terminal city for a tour. He gains happiness (*H*<sub class="lower-index">*i*</sub> - *L*) during a tour, where *L* is the distance between the city *A*<sub class="lower-index">*i*</sub> and the terminal city.

Ralph is interested in tours from *A*<sub class="lower-index">*i*</sub> in which he can gain positive happiness. For each query, compute the sum of happiness gains for all such tours.

Ralph will never take the same tour twice or more (in one query), he will never pass the same city twice or more in one tour.
The first line contains two integers *n* and *m* (1 ≤ *n* ≤ 10<sup class="upper-index">6</sup>, 1 ≤ *m* ≤ 10<sup class="upper-index">5</sup>).

(*n* - 1) lines follow, each line contains one integer *L*<sub class="lower-index">*i*</sub> (1 ≤ *L*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>), which denotes the length of the *i*-th road.

*m* lines follow, each line contains two integers *A*<sub class="lower-index">*i*</sub> and *H*<sub class="lower-index">*i*</sub> (1 ≤ *A*<sub class="lower-index">*i*</sub> ≤ *n*, 0 ≤ *H*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">7</sup>).
Here is the explanation for the second sample.

Ralph's first query is to start tours from city 2 and *H*<sub class="lower-index">*i*</sub> equals to 4. Here are the options:
 -  He can choose city 5 as his terminal city. Since the distance between city 5 and city 2 is 3, he can gain happiness 4 - 3 = 1. -  He can choose city 4 as his terminal city and gain happiness 3. -  He can choose city 1 as his terminal city and gain happiness 2. -  He can choose city 3 as his terminal city and gain happiness 1. -  Note that Ralph can choose city 2 as his terminal city and gain happiness 4. -  Ralph won't choose city 6 as his terminal city because the distance between city 6 and city 2 is 5, which leads to negative happiness for Ralph. 
So the answer for the first query is 1 + 3 + 2 + 1 + 4 = 11.
2 2
5
1 8
2 4
11
4
6 4
2
1
1
3
2
2 4
1 3
3 2
1 7
11
6
3
28
8 1
21725
80273
97276
78838
78474
1896
6570
7 5267977
41283845
4 2
56025
27554
51024
2 4730374
4 7372300
18730868
29196524
7 3
73302
43581
59091
35794
64662
60029
6 4980178
4 2677480
5 9024109
33924130
17803378
62346266
1 1
1 1
1
for s in[*open(0)][2::2]:b=min(a:=[x&-x for x in
map(int,s.split())]);print(len(a)+min(k:=bin(b)[::-1].find('1')-1,k*a.count(b)))