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;
int const N = 200000, md = 1e9 + 7;
int a[N + 1], n, q, aw[N + 1];
long long fn[N + 1];
void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
void sb(int &x, int y) {
  if ((x -= y) < 0) x += md;
}
void adf(int i, int x) {
  while (i <= n) fn[i] += x, i += i & -i;
}
long long gt(int i) {
  long long an = 0;
  while (i) an += fn[i], i ^= i & -i;
  return an;
}
long long gt(int l, int r) { return gt(r) - gt(l - 1); }
int bs(long long v) {
  int j = 0;
  for (int i = 17; i >= 0; --i) {
    int nj = j + (1 << i);
    if (nj > n) continue;
    if (fn[nj] <= v) {
      j = nj;
      v -= fn[nj];
    }
  }
  return j;
}
void adw(int i, int x) {
  while (i <= n) {
    if (x < 0)
      sb(aw[i], -x);
    else
      ad(aw[i], x);
    i += i & -i;
  }
}
int gtw(int i) {
  int an = 0;
  while (i) ad(an, aw[i]), i ^= i & -i;
  return an;
}
int gtw(int l, int r) {
  int an = gtw(r);
  sb(an, gtw(l - 1));
  return an;
}
void an(int l, int r, int m) {
  int ds = a[m];
  int o = 0;
  sb(o, gtw(l, m));
  ad(o, gt(l, m) % md * ds % md);
  ad(o, gtw(m + 1, r));
  sb(o, gt(m + 1, r) % md * ds % md);
  printf("%d\n", o);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < (int)(n + 1); ++i) scanf("%d", a + i);
  for (int i = 1; i < (int)(n + 1); ++i) a[i] -= i;
  for (int i = 1; i < (int)(n + 1); ++i) {
    int t;
    scanf("%d", &t);
    adw(i, (long long)a[i] * t % md);
    adf(i, t);
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (l < 0) {
      l = -l;
      int t = (long long)a[l] * r % md;
      sb(t, gtw(l, l));
      adw(l, t);
      adf(l, r - gt(l, l));
    } else {
      if (l == r) {
        printf("0\n");
        continue;
      }
      long long sg = gt(l, r) >> 1;
      int i = bs(sg + gt(l - 1));
      long long a = gt(l, i), b = gt(i + 1, r);
      if (a > b)
        an(l, r, i);
      else
        an(l, r, i + 1);
    }
  }
}