Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.Arrays;

public class Main {
    private static final int c = (int) (2e5 + 10), mod = (int) (1e9 + 7);
    static int n, q;
    static long[] a = new long[c];
    static long[] w = new long[c];
    static long[][] tre = new long[c][2];

    static void update(int i, long x, int j) {
        while (i <= n) {
            tre[i][j] += x;
            if (j == 1) tre[i][j] %= mod;
            i += i & -i;
        }
    }

    static long query(int i, int j) {
        long s = 0;
        while (i > 0) {
            s += tre[i][j];
            if (j == 1) s %= mod;
            i -= i & -i;
        }
        return s;
    }

    public static void main(String[] args) {
        IO io = new IO();
        n = io.nextInt();
        q = io.nextInt();
        for (int i = 0; i < n; i++) a[i] = io.nextLong();
        for (int i = 0; i < n; i++) {
            update(i + 1, w[i] = io.nextLong(), 0);
            update(i + 1, w[i] * (a[i] - i), 1);
        }
        while (q-- > 0) {
            int x = io.nextInt(), y = io.nextInt();
            if (x < 0) {
                x = -x - 1;
                update(x + 1, y - w[x], 0);
                update(x + 1, (y - w[x]) * (a[x] - x), 1);
                w[x] = y;
            } else {
                long s = query(y, 0) + query(x - 1, 0), c = 0;
                int mid = 0;
                for (int i = 17; i >= 0; i--)
                    if (mid + (1 << i) < n && (c + tre[mid + (1 << i)][0]) * 2 < s) {
                        mid += 1 << i;
                        c += tre[mid][0];
                    }
                long a1 = query(y, 1) - 2 * query(mid, 1) + query(x-1, 1);
                long a2 = query(y, 0) - 2 * query(mid, 0) + query(x-1, 0);
                long ans = a1 % mod - a2 % mod * (a[mid] - mid) % mod + 2 * mod;
                io.println(ans % mod );

            }
        }
    }

    static class IO {

        BufferedInputStream din;
        final int BUFFER_SIZE = 1 << 16;
        byte[] buffer;
        int byteRead, bufferPoint;

        StringBuilder builder;
        PrintWriter pw;

        public IO() {
            din = new BufferedInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPoint = byteRead = 0;

            builder = new StringBuilder();
            pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(
                    System.out
            )), true);
        }

        int read() {
            if (bufferPoint >= byteRead) {
                try {
                    byteRead = din.read(buffer, bufferPoint = 0, BUFFER_SIZE);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (byteRead == -1) buffer[0] = -1;
            }
            return buffer[bufferPoint++];
        }

        int peek() {
            if (byteRead == -1) return -1;
            if (bufferPoint >= byteRead) {
                try {
                    byteRead = din.read(buffer, bufferPoint = 0, BUFFER_SIZE);
                } catch (IOException e) {
                    return -1;
                }
                if (byteRead <= 0) return -1;
            }
            return buffer[bufferPoint];
        }

        boolean hasNext() {
            int c = peek();
            while (c != -1 && c <= ' ') {
                read();
                c = peek();
            }
            return c != -1;
        }

        char nextChar() {
            int c = read();
            while (c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1) {
                c = read();
            }
            return (char) c;
        }

        double nextDouble() {
            double ret = 0, div = 1;
            int c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
            if (neg)
                return -ret;
            return ret;
        }

        String nextLine() {
            byte[] strBuf = new byte[64];
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt == 0) {
                        continue;
                    } else {
                        break;
                    }
                }
                if (strBuf.length == cnt) {
                    strBuf = Arrays.copyOf(strBuf, strBuf.length * 2);
                }
                strBuf[cnt++] = (byte) c;
            }
            return new String(strBuf, 0, cnt);
        }


        String next() {
            byte[] strBuf = new byte[64];
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (Character.isWhitespace(c)) {
                    if (cnt == 0) {
                        continue;
                    } else {
                        break;
                    }
                }
                if (strBuf.length == cnt) {
                    strBuf = Arrays.copyOf(strBuf, strBuf.length * 2);
                }
                strBuf[cnt++] = (byte) c;
            }
            return new String(strBuf, 0, cnt);
        }

        int nextInt() {
            int ans = 0;
            int c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ans = ans * 10 + c - '0';
            } while ('0' <= (c = read()) && c <= '9');
            bufferPoint--;
            return neg ? -ans : ans;
        }

        long nextLong() {
            long ans = 0;
            int c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ans = ans * 10 + c - '0';
            } while ('0' <= (c = read()) && c <= '9');
            bufferPoint--;
            return neg ? -ans : ans;
        }

        void println(Object o) {
            pw.println(o);
        }

        void print(Object o) {
            pw.print(o);
        }

        void printf(String format, Object... objects) {
            pw.printf(format, objects);
        }

        void close() {
            pw.close();
        }

        void done(Object o) {
            print(o);
            close();
        }

    }

}