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.*;
import java.math.BigInteger;
import java.util.*;

public class Solution {

    public static void main(String[] args) throws Exception {
        MyReader reader = new MyReader(System.in);
        MyWriter writer = new MyWriter(System.out);
        new Solution().run(reader, writer);
        writer.close();
    }

    private void run(MyReader reader, MyWriter writer) throws IOException {
        int n = reader.nextInt();
        int m = reader.nextInt();
        int[] l = new int[n];
        for (int i = 1; i < n; i++) {
            l[i] = reader.nextInt();
        }
        long[][] d = new long[n + 1][0];
        long[][] p = new long[n + 1][0];
        for (int i = n; i > 0; i--) {
            if (i > n / 2) {
                d[i] = new long[1];
                d[i][0] = 0;
            } else {
                long[] a1 = i * 2 > n ? d[0] : d[i * 2];
                long[] a2 = i * 2 + 1 > n ? d[0] : d[i * 2 + 1];
                long[] a3 = new long[a1.length + a2.length + 1];
                int i1 = 0;
                int i2 = 0;
                int i3 = 1;
                while (i1 < a1.length || i2 < a2.length) {
                    if (i1 == a1.length) {
                        a3[i3++] = a2[i2++] + l[i * 2];
                    } else if (i2 == a2.length) {
                        a3[i3++] = a1[i1++] + l[i * 2 - 1];
                    } else if (a1[i1] + l[i * 2 - 1] > a2[i2] + l[i * 2]) {
                        a3[i3++] = a2[i2++] + l[i * 2];
                    } else {
                        a3[i3++] = a1[i1++] + l[i * 2 - 1];
                    }
                }
                d[i] = a3;
            }
        }
        for (int i = 1; i <= n; i++) {
            p[i] = new long[d[i].length];
            for (int j = 1; j < p[i].length; j++) {
                p[i][j] = p[i][j - 1] + d[i][j];
            }
        }
        for (int i = 0; i < m; i++) {
            int ai = reader.nextInt();
            long hi = reader.nextInt();
            long ans = f(ai, hi, d, p);
            while (ai > 1) {
                if (l[ai - 1] >= hi) {
                    break;
                }
                hi -= l[ai - 1];
                ans += hi;
                int ai2;
                if (ai % 2 == 0) {
                    ai2 = ai + 1;
                } else {
                    ai2 = ai - 1;
                }
                ans += ai2 > n ? 0 : f(ai2, hi - l[ai2 - 1], d, p);
                ai >>= 1;
            }
            writer.println(ans + "");
        }
    }

    private long f(int ai, long hi, long[][] d, long[][] p) {
        int a = Arrays.binarySearch(d[ai], hi);
        if (a < 0) {
            a = -a - 1;
        }
        return a > 0 ? hi * a - p[ai][a - 1] : 0;
    }

    static class MyReader {

        final BufferedInputStream in;
        final int bufSize = 1 << 16;
        final byte buf[] = new byte[bufSize];
        int i = bufSize;
        int k = bufSize;
        final StringBuilder str = new StringBuilder();

        MyReader(InputStream in) {
            this.in = new BufferedInputStream(in, bufSize);
        }

        int nextInt() throws IOException {
            return (int) nextLong();
        }

        int[] nextIntArray(int n) throws IOException {
            int[] m = new int[n];
            for (int i = 0; i < n; i++) {
                m[i] = nextInt();
            }
            return m;
        }

        int[][] nextIntMatrix(int n, int m) throws IOException {
            int[][] a = new int[n][0];
            for (int j = 0; j < n; j++) {
                a[j] = nextIntArray(m);
            }
            return a;
        }

        long nextLong() throws IOException {
            int c;
            long x = 0;
            boolean sign = true;
            while ((c = nextChar()) <= 32) ;
            if (c == '-') {
                sign = false;
                c = nextChar();
            }
            if (c == '+') {
                c = nextChar();
            }
            while (c >= '0') {
                x = x * 10 + (c - '0');
                c = nextChar();
            }
            return sign ? x : -x;
        }

        long[] nextLongArray(int n) throws IOException {
            long[] m = new long[n];
            for (int i = 0; i < n; i++) {
                m[i] = nextLong();
            }
            return m;
        }

        int nextChar() throws IOException {
            if (i == k) {
                k = in.read(buf, 0, bufSize);
                i = 0;
            }
            return i >= k ? -1 : buf[i++];
        }

        String nextString() throws IOException {
            int c;
            str.setLength(0);
            while ((c = nextChar()) <= 32 && c != -1) ;
            if (c == -1) {
                return null;
            }
            while (c > 32) {
                str.append((char) c);
                c = nextChar();
            }
            return str.toString();
        }

        String nextLine() throws IOException {
            int c;
            str.setLength(0);
            while ((c = nextChar()) <= 32 && c != -1) ;
            if (c == -1) {
                return null;
            }
            while (c != '\n') {
                str.append((char) c);
                c = nextChar();
            }
            return str.toString();
        }

        char[] nextCharArray() throws IOException {
            return nextString().toCharArray();
        }

        char[][] nextCharMatrix(int n) throws IOException {
            char[][] a = new char[n][0];
            for (int i = 0; i < n; i++) {
                a[i] = nextCharArray();
            }
            return a;
        }
    }

    static class MyWriter {

        final BufferedOutputStream out;
        final int bufSize = 1 << 16;
        final byte buf[] = new byte[bufSize];
        int i = 0;
        final byte c[] = new byte[30];
        static final String newLine = System.getProperty("line.separator");

        MyWriter(OutputStream out) {
            this.out = new BufferedOutputStream(out, bufSize);
        }

        void print(long x) throws IOException {
            int j = 0;
            if (i + 30 >= bufSize) {
                flush();
            }
            if (x < 0) {
                buf[i++] = (byte) ('-');
                x = -x;
            }
            while (j == 0 || x != 0) {
                c[j++] = (byte) (x % 10 + '0');
                x /= 10;
            }
            while (j-- > 0)
                buf[i++] = c[j];
        }

        void print(int[] m) throws IOException {
            for (int a : m) {
                print(a);
                print(' ');
            }
        }

        void print(long[] m) throws IOException {
            for (long a : m) {
                print(a);
                print(' ');
            }
        }

        void print(String s) throws IOException {
            for (int i = 0; i < s.length(); i++) {
                print(s.charAt(i));
            }
        }

        void print(char x) throws IOException {
            if (i == bufSize) {
                flush();
            }
            buf[i++] = (byte) x;
        }

        void print(char[] m) throws IOException {
            for (char c : m) {
                print(c);
            }
        }

        void println(String s) throws IOException {
            print(s);
            println();
        }

        void println() throws IOException {
            print(newLine);
        }

        void flush() throws IOException {
            out.write(buf, 0, i);
            out.flush();
            i = 0;
        }

        void close() throws IOException {
            flush();
            out.close();
        }
    }
}