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>
const int mod = 1000000007;
struct Segment_Tree {
  long long val, s;
} tree[1000100];
int w[200100], a[200100], n, Q, x, y;
void build(int t, int l, int r) {
  if (l == r) {
    tree[t].val = w[l];
    tree[t].s = 1ll * a[l] * w[l] % mod;
    return;
  }
  int mid = (l + r) / 2;
  build(t * 2, l, mid);
  build(t * 2 + 1, mid + 1, r);
  tree[t].val = tree[t * 2].val + tree[t * 2 + 1].val;
  tree[t].s = (tree[t * 2].s + tree[t * 2 + 1].s) % mod;
}
void modify(int t, int l, int r, int pos, int val) {
  if (l == r) {
    w[l] = val;
    tree[t].val = w[l];
    tree[t].s = 1ll * a[l] * w[l] % mod;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    modify(t * 2, l, mid, pos, val);
  else
    modify(t * 2 + 1, mid + 1, r, pos, val);
  tree[t].val = tree[t * 2].val + tree[t * 2 + 1].val;
  tree[t].s = (tree[t * 2].s + tree[t * 2 + 1].s) % mod;
}
long long queryval(int t, int l, int r, int L, int R) {
  if (L > r || l > R) return 0ll;
  if (L <= l && r <= R) return tree[t].val;
  int mid = (l + r) / 2;
  return queryval(t * 2, l, mid, L, R) + queryval(t * 2 + 1, mid + 1, r, L, R);
}
long long querys(int t, int l, int r, int L, int R) {
  if (L > r || l > R) return 0ll;
  if (L <= l && r <= R) return tree[t].s;
  int mid = (l + r) / 2;
  return (querys(t * 2, l, mid, L, R) + querys(t * 2 + 1, mid + 1, r, L, R)) %
         mod;
}
void solve(int l, int r) {
  int L = l, R = r, pos;
  while (L <= R) {
    int mid = (L + R) / 2;
    if (queryval(1, 1, n, l, mid) >= queryval(1, 1, n, mid + 1, r))
      R = mid - 1, pos = mid;
    else
      L = mid + 1;
  }
  long long res = 0;
  res = (-1ll * querys(1, 1, n, l, pos) +
         queryval(1, 1, n, l, pos) % mod * a[pos] % mod + mod) %
        mod;
  res = (res - queryval(1, 1, n, pos, r) % mod * a[pos] % mod +
         querys(1, 1, n, pos, r) + mod) %
        mod;
  printf("%lld\n", res);
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] -= i;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  build(1, 1, n);
  while (Q--) {
    scanf("%d%d", &x, &y);
    if (x < 0)
      modify(1, 1, n, -x, y);
    else
      solve(x, y);
  }
}