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;
void add(long long &a, long long b) {
  a += b;
  if (a > 1000000007) a -= 1000000007;
}
int N, Q, a[200200];
long long w[200200];
struct modtree {
  long long t[200200];
  void u(int i, long long v) {
    i++;
    v %= 1000000007;
    for (; i <= N; i += (i & -i)) add(t[i], v);
  }
  long long gtv(int i) {
    i++;
    long long res = 0;
    for (; i > 0; i -= (i & -i)) add(res, t[i]);
    return res;
  }
  long long q(int l, int r) {
    return (gtv(r) - gtv(l - 1) + 1000000007) % 1000000007;
  }
} awt, wit;
struct tree {
  long long t[200200];
  void u(int i, long long v) {
    i++;
    for (; i <= N; i += (i & -i)) t[i] += v;
  }
  long long gtv(int i) {
    i++;
    long long res = 0;
    for (; i > 0; i -= (i & -i)) res += t[i];
    return res;
  }
  long long q(int l, int r) { return gtv(r) - gtv(l - 1); }
} wt;
long long dif(int l, int r, int s) {
  if (l > r) return 0ll;
  long long wi_tot = wit.q(l, r);
  long long wt_tot = wt.q(l, r) % 1000000007;
  wi_tot = (wi_tot + wt_tot * s) % 1000000007;
  long long aw_tot = awt.q(l, r);
  return (wi_tot - aw_tot + 1000000007) % 1000000007;
}
long long delta(int l, int r, int s) {
  int lo = l, hi = r, split = l - 1;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (a[mid] <= s + mid)
      lo = mid + 1, split = mid;
    else
      hi = mid - 1;
  }
  long long d = 0;
  if (l <= split) d += wt.q(l, split);
  if (split + 1 <= r) d -= wt.q(split + 1, r);
  return d;
}
long long dis(int l, int r, int s) {
  int lo = l, hi = r, split = l - 1;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (a[mid] <= s + mid)
      lo = mid + 1, split = mid;
    else
      hi = mid - 1;
  }
  long long lft = dif(l, split, s);
  long long rgt = dif(split + 1, r, s);
  return (lft - rgt + 1000000007) % 1000000007;
}
int main() {
  scanf("%d %d", &N, &Q);
  for (int i = (0); i < (N); i++) scanf("%d", &a[i]);
  for (int i = (0); i < (N); i++) {
    scanf("%lld", &w[i]);
    wt.u(i, w[i]);
    awt.u(i, w[i] * a[i] % 1000000007);
    wit.u(i, w[i] * i % 1000000007);
  }
  for (int i = (0); i < (Q); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      x = -x - 1;
      wt.u(x, y - w[x]);
      long long dif = ((long long)y - w[x] + 1000000007) % 1000000007;
      w[x] = (long long)y;
      awt.u(x, dif * a[x] % 1000000007);
      wit.u(x, dif * x % 1000000007);
    } else {
      x--, y--;
      int lo = 1, hi = (int)1e9, ans = -1;
      while (lo <= hi) {
        int mid = (lo + hi) / 2;
        long long res = delta(x, y, mid);
        if (res >= 0) {
          hi = mid - 1;
          ans = mid;
        } else
          lo = mid + 1;
      }
      long long u = dis(x, y, ans);
      printf("%lld\n", u);
    }
  }
  return 0;
}