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;
const int mod = 1e9 + 7;
int add(int x, int y) { return ((x + y) % mod + mod) % mod; }
int mul(int x, int y) { return (long long)x * y % mod; }
int mypow(int x, int c) {
  int ret = 1;
  while (c > 0) {
    if (c & 1) {
      ret = mul(ret, x);
    }
    c /= 2;
    x = mul(x, x);
  }
  return ret;
}
long long pen[200005];
void update(int ind, int add) {
  while (ind < 200005) {
    pen[ind] += add;
    ind += ind & (-ind);
  }
}
long long query(int ind) {
  long long ret = 0;
  while (ind > 0) {
    ret = ret + pen[ind];
    ind = ind & (ind - 1);
  }
  return ret;
}
int a[200005], w[200005];
int pen2[200005];
int pen3[200005];
void update2(int *pp, int ind, int ad) {
  while (ind < 200005) {
    pp[ind] = add(pp[ind], ad);
    ind += ind & (-ind);
  }
}
int query2(int *pp, int ind) {
  int ret = 0;
  while (ind > 0) {
    ret = add(ret, pp[ind]);
    ind = ind & (ind - 1);
  }
  return ret;
}
int getS(int x, int y, int wei) {
  long long s = query(y) - query(x - 1);
  int sum = s % mod;
  int r = add(add(query2(pen3, y), -query2(pen3, x - 1)), -mul(sum, x));
  return add(r, mul(sum, wei));
}
void solve() {
  int n, q;
  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]);
    update(i, w[i]);
    update2(pen2, i, mul(w[i], a[i]));
    update2(pen3, i, mul(i, w[i]));
  }
  while (q--) {
    int x, y;
    scanf("%d %d ", &x, &y);
    if (x < 0) {
      int id = -x;
      update(id, -w[id]);
      update2(pen2, id, -mul(w[id], a[id]));
      update2(pen3, id, -mul(id, w[id]));
      w[id] = y;
      update(id, w[id]);
      update2(pen2, id, mul(w[id], a[id]));
      update2(pen3, id, mul(id, w[id]));
    } else {
      long long pre = query(x - 1);
      long long sum = query(y) - pre;
      int l = x, r = y, mid;
      while (l < r) {
        mid = (l + r) / 2;
        if ((query(mid) - pre) * 2 > sum) {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      int ind = l;
      int ret = 0;
      if (x <= ind - 1) {
        int wei = a[ind] - ind + x;
        ret = add(ret, add(getS(x, ind - 1, wei),
                           -add(query2(pen2, ind - 1), -query2(pen2, x - 1))));
      }
      if (ind + 1 <= y) {
        int wei = a[ind] + 1;
        ret = add(ret, add(-getS(ind + 1, y, wei),
                           add(query2(pen2, y), -query2(pen2, ind))));
      }
      printf("%d\n", ret);
    }
  }
}
int main() {
  solve();
  return 0;
}