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 mo = 1000000007;
const int N = 200005;
int n, q;
int a[N], w[N];
long long cntw[N * 4];
int val[N * 4];
void build(int k, int l, int r) {
  if (l == r) {
    cntw[k] = w[l];
    val[k] = 1ll * w[l] * a[l] % mo;
    return;
  }
  int mid = (l + r) / 2;
  build(k * 2, l, mid);
  build(k * 2 + 1, mid + 1, r);
  cntw[k] = cntw[k * 2] + cntw[k * 2 + 1];
  val[k] = (val[k * 2] + val[k * 2 + 1]) % mo;
}
void change(int k, int l, int r, int p, int v) {
  if (l == r) {
    cntw[k] = v;
    val[k] = 1ll * v * a[l] % mo;
    return;
  }
  int mid = (l + r) / 2;
  if (p <= mid)
    change(k * 2, l, mid, p, v);
  else
    change(k * 2 + 1, mid + 1, r, p, v);
  cntw[k] = cntw[k * 2] + cntw[k * 2 + 1];
  val[k] = (val[k * 2] + val[k * 2 + 1]) % mo;
}
long long askw(int k, int l, int r, int x, int y) {
  if (l == x && r == y) return cntw[k];
  int mid = (l + r) / 2;
  if (y <= mid) return askw(k * 2, l, mid, x, y);
  if (x > mid) return askw(k * 2 + 1, mid + 1, r, x, y);
  return askw(k * 2, l, mid, x, mid) + askw(k * 2 + 1, mid + 1, r, mid + 1, y);
}
int findpos(int k, int l, int r, long long v) {
  if (l == r) return l;
  int mid = (l + r) / 2;
  if (cntw[k * 2] <= v) return findpos(k * 2 + 1, mid + 1, r, v - cntw[k * 2]);
  return findpos(k * 2, l, mid, v);
}
int askcnt(int k, int l, int r, int x, int y) {
  if (l == x && r == y) return val[k];
  int mid = (l + r) / 2;
  if (y <= mid) return askcnt(k * 2, l, mid, x, y);
  if (x > mid) return askcnt(k * 2 + 1, mid + 1, r, x, y);
  return (askcnt(k * 2, l, mid, x, mid) +
          askcnt(k * 2 + 1, mid + 1, r, mid + 1, y)) %
         mo;
}
int calc(int l, int mid, int r) {
  int ans = 0;
  if (mid >= l) {
    ans = mo - askcnt(1, 1, n, l, mid);
    ans = (ans + 1ll * askw(1, 1, n, l, mid) % mo * a[mid] % mo) % mo;
  }
  if (mid < r) {
    ans = (ans + askcnt(1, 1, n, mid + 1, r)) % mo;
    ans = (ans + mo - 1ll * askw(1, 1, n, mid + 1, r) % mo * a[mid] % mo) % mo;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] += n - i;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  build(1, 1, n);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0)
      change(1, 1, n, -x, y);
    else {
      long long sum = askw(1, 1, n, x, y);
      long long leftv = (sum + 1) / 2;
      if (x != 1) leftv += askw(1, 1, n, 1, x - 1);
      int mid = findpos(1, 1, n, leftv - 1);
      printf("%d\n", calc(x, mid, y));
    }
  }
}