Problem: 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 <image> (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 Li.

Now Ralph gives you m queries. In each query he tells you some city Ai and an integer Hi. He wants to make some tours starting from this city. He can choose any city in the Binary Country (including Ai) as the terminal city for a tour. He gains happiness (Hi - L) during a tour, where L is the distance between the city Ai and the terminal city.

Ralph is interested in tours from Ai 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.

Input

The first line contains two integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 105).

(n - 1) lines follow, each line contains one integer Li (1 ≤ Li ≤ 105), which denotes the length of the i-th road.

m lines follow, each line contains two integers Ai and Hi (1 ≤ Ai ≤ n, 0 ≤ Hi ≤ 107).

Output

Print m lines, on the i-th line print one integer — the answer for the i-th query.

Examples

Input

2 2
5
1 8
2 4


Output

11
4


Input

6 4
2
1
1
3
2
2 4
1 3
3 2
1 7


Output

11
6
3
28

Note

Here is the explanation for the second sample.

Ralph's first query is to start tours from city 2 and Hi 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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class D {
	
	static int N;
	static int[][] cost;
	static int[][] dec;
	static long[][] preSum;
	
	public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		N = sc.nextInt();
		int m = sc.nextInt();
		cost = new int[2][N + 1];
		dec = new int[N + 1][0];
		preSum = new long[N + 1][0];
		
		int len = N >> 1;
		for(int i = 1; i <= len; i++)
		{
			cost[0][i] = sc.nextInt();
			if(i == len && (N & 1) == 0) break;
			cost[1][i] = sc.nextInt();
		}
		dfs(1);
		
		for(int i = 0; i < m; i++)
		{
			int node = sc.nextInt(), h = sc.nextInt();
			out.println(solve(node, h));
		}
		
		out.flush();
		out.close();
	}
	
	static long solve(int node, int h)
	{
		int idx = bs(node, h);
		long sum = idx == -1? 0 : preSum[node][idx];
		int count = idx + 2;
		int dir = (node & 1);
		int tale3Sum = 0;
		while(true)
		{
			node >>= 1;
			if(node < 1 || tale3Sum + cost[dir][node] > h) break;
			tale3Sum += cost[dir][node];
			count++;
			sum += tale3Sum;
			int costTale3Nazel = tale3Sum + cost[dir ^ 1][node];
			int nodeTale3Nazel = (node << 1) | (dir ^ 1);
			if(costTale3Nazel <= h && nodeTale3Nazel <= N)
			{
				idx = bs(nodeTale3Nazel, h - costTale3Nazel);
				sum += 1l * costTale3Nazel * (idx + 2) + (idx == -1? 0 : preSum[nodeTale3Nazel][idx]);
				count += idx + 2;
			}
			dir = (node & 1);
		}
		return 1l * h * count - sum;
	}
	
	static int bs(int idx, int h)
	{
		int lo = 0, hi = dec[idx].length-1, ans = -1;
		while(lo <= hi)
		{
			int mid = (lo + hi) >> 1;
			if(dec[idx][mid] <= h)
				lo = (ans = mid) + 1;
			else 
				hi = mid - 1;
		}
		return ans;
	}
	
	static void dfs(int u)
	{
		int l = u << 1, r = (u << 1) | 1;
		if(l > N) return;
		dfs(l); dfs(r);

		int[] decL = new int[dec[l].length + 1];
		decL[0] = cost[0][u];
		for(int i = 0; i < dec[l].length; i++) decL[i + 1] = cost[0][u] + dec[l][i];
		int[] decR = new int[0];
		if(r <= N)
		{
			decR = new int[dec[r].length + 1];
			decR[0] = cost[1][u];
			for(int i = 0; i < dec[r].length; i++) decR[i + 1] = cost[1][u] + dec[r][i];
		}
		
		dec[u] = new int[decR.length + decL.length];
		preSum[u] = new long[decR.length + decL.length];
		
		int i = 0, j = 0, idx = 0;
		while(i < decL.length || j < decR.length)
		{
			int cur;

			if(j == decR.length)
				dec[u][idx] = (cur = decL[i++]);
			else if(i == decL.length)
				dec[u][idx] = (cur = decR[j++]);
			else if(decL[i] < decR[j])
				dec[u][idx] = (cur = decL[i++]);
			else
				dec[u][idx] = (cur = decR[j++]);
			
			if(idx == 0) preSum[u][idx] = (long)cur;
			else preSum[u][idx] = cur + preSum[u][idx - 1];
			idx++;
		}
	}
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}