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 n, q;
const int N = 2e5 + 10;
const int Mod = 1e9 + 7;
void inc(int &x, int y) {
  x += y;
  x = x >= Mod ? x - Mod : x;
}
struct Binary_Indexed_Tree {
  long long c[N];
  int lowbit(int x) { return x & (-x); }
  void add(int x, int w) {
    while (x <= n) {
      c[x] += w;
      x += lowbit(x);
    }
  }
  long long sum(int x) {
    long long ret = 0;
    while (x) {
      ret += c[x];
      x -= lowbit(x);
    }
    return ret;
  }
} T, S;
int w[N], a[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 1; i <= n; i++)
    T.add(i, w[i]), S.add(i, 1ll * (a[i] - i) * w[i] % Mod);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      T.add(x, y - w[x]);
      S.add(x, 1ll * (a[x] - x) * (y - w[x] + Mod) % Mod);
      w[x] = y;
    } else {
      long long lw = T.sum(x - 1), sumw = T.sum(y) - lw;
      int l = x, r = y;
      while (l < r) {
        int mid = (l + r) / 2;
        if ((T.sum(mid) - lw) * 2 >= sumw)
          r = mid;
        else
          l = mid + 1;
      }
      long long ans = S.sum(x - 1) + S.sum(y) - 2ll * S.sum(l);
      ans = ans + (2ll * T.sum(l) - T.sum(x - 1) - T.sum(y)) % Mod * (a[l] - l);
      ans = (ans % Mod + Mod) % Mod;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}