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 P = 1000000007, N = 200005;
int n, m;
int a[N], w[N], caw[4 * N], Ans;
long long cw[4 * N], Sum;
long long queryw(int p, int l, int r, int x, int y) {
  if (x > y) return 0;
  if (l == x && r == y) return cw[p];
  int mid = (l + r) >> 1, ll = p << 1;
  if (y <= mid)
    return queryw(ll, l, mid, x, y);
  else if (x >= mid + 1)
    return queryw(ll | 1, mid + 1, r, x, y);
  else
    return queryw(ll, l, mid, x, mid) + queryw(ll | 1, mid + 1, r, mid + 1, y);
}
int queryaw(int p, int l, int r, int x, int y) {
  if (x > y) return 0;
  if (l == x && r == y) return caw[p];
  int mid = (l + r) >> 1, ll = p << 1;
  if (y <= mid)
    return queryaw(ll, l, mid, x, y);
  else if (x >= mid + 1)
    return queryaw(ll | 1, mid + 1, r, x, y);
  else
    return (queryaw(ll, l, mid, x, mid) +
            queryaw(ll | 1, mid + 1, r, mid + 1, y)) %
           P;
}
void build(int p, int l, int r) {
  if (l == r) {
    caw[p] = 1ll * (a[l] + P - l) * w[l] % P;
    cw[p] = w[l];
    return;
  }
  int mid = (l + r) >> 1, ll = p << 1;
  build(ll, l, mid);
  build(ll | 1, mid + 1, r);
  cw[p] = cw[ll] + cw[ll | 1];
  caw[p] = (caw[ll] + caw[ll | 1]) % P;
}
void change(int p, int l, int r, int x) {
  if (l == r) {
    caw[p] = 1ll * (a[l] + P - l) * w[l] % P;
    cw[p] = w[l];
    return;
  }
  int mid = (l + r) >> 1, ll = p << 1;
  if (x <= mid)
    change(ll, l, mid, x);
  else
    change(ll | 1, mid + 1, r, x);
  cw[p] = cw[ll] + cw[ll | 1];
  caw[p] = (caw[ll] + caw[ll | 1]) % P;
}
int findKth(int p, int l, int r) {
  if (l == r) return l;
  int mid = (l + r) >> 1, ll = p << 1;
  if (cw[ll] >= Sum - cw[ll])
    return findKth(ll, l, mid);
  else {
    Sum -= 2 * cw[ll];
    return findKth(ll | 1, mid + 1, r);
  }
}
int find(int p, int l, int r, int x, int y) {
  if (l == x && r == y) {
    if (cw[p] >= Sum - cw[p])
      return findKth(p, l, r);
    else {
      Sum -= 2 * cw[p];
      return -1;
    }
  }
  int mid = (l + r) >> 1, ll = p << 1;
  if (y <= mid)
    return find(ll, l, mid, x, y);
  else if (x >= mid + 1)
    return find(ll | 1, mid + 1, r, x, y);
  else {
    int tmp = find(ll, l, mid, x, mid);
    if (tmp != -1) return tmp;
    return find(ll | 1, mid + 1, r, mid + 1, y);
  }
}
void upd(int &x, long long y) { x = (x + y) % P; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &w[i]);
  build(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      w[-x] = y;
      change(1, 1, n, -x);
    } else {
      Sum = queryw(1, 1, n, x, y);
      int K = find(1, 1, n, x, y);
      Ans = 0;
      upd(Ans, 1ll * (a[K] + P - K) * (queryw(1, 1, n, x, K) % P));
      upd(Ans, P - queryaw(1, 1, n, x, K));
      upd(Ans, 1ll * (K + P - a[K]) * (queryw(1, 1, n, K + 1, y) % P));
      upd(Ans, queryaw(1, 1, n, K + 1, y));
      printf("%d\n", Ans);
    }
  }
  return 0;
}