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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  char c;
  while ((c = getchar()) < '-')
    ;
  if (c == '-') {
    int x = (c = getchar()) - '0';
    while ((c = getchar()) >= '0') x = x * 10 + c - '0';
    return -x;
  }
  int x = c - '0';
  while ((c = getchar()) >= '0') x = x * 10 + c - '0';
  return x;
}
const int N = 2e5 + 5, T = N * 3, D = 1e9 + 7;
int b[N];
long long s[T], s2[T];
int n, d;
void modify(int i, int w) {
  i += d;
  s[i] = w;
  s2[i] = (long long)w * b[i - d] % D;
  while (i >>= 1) {
    s[i] = s[(i * 2)] + s[((i * 2) + 1)];
    s2[i] = s2[(i * 2)] + s2[((i * 2) + 1)];
  }
}
long long qiu(int i, int t) {
  if (i == 0) return 0;
  i += d;
  long long a1 = s[i], a2 = s2[i];
  while (i) {
    if (i % 2) {
      a1 += s[i - 1];
      a2 += s2[i - 1];
    }
    i >>= 1;
  }
  return t == 1 ? a1 : a2;
}
int erfen(long long sum) {
  int i = 1;
  while (i <= d)
    if (s[(i * 2)] >= sum)
      i = (i * 2);
    else {
      sum -= s[(i * 2)];
      i = ((i * 2) + 1);
    }
  return i - d;
}
int query(int l, int r) {
  long long sum = qiu(r, 1) - qiu(l - 1, 1);
  int m = erfen((sum + 1) / 2 + qiu(l - 1, 1));
  return ((b[m] * ((qiu(m, 1) - qiu(l - 1, 1) - (qiu(r, 1) - qiu(m, 1))) % D) +
           (qiu(r, 2) - qiu(m, 2) - (qiu(m, 2) - qiu(l - 1, 2)))) %
              D +
          D) %
         D;
}
int main() {
  n = read();
  int m = read();
  for (d = 1; d < n; d *= 2)
    ;
  --d;
  for (int i = 1; i <= n; ++i) b[i] = read() - i;
  for (int i = 1; i <= n; ++i) modify(i, read());
  while (m--) {
    int x = read(), y = read();
    if (x < 0)
      modify(-x, y);
    else
      printf("%d\n", query(x, y));
  }
}