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;
template <typename T1, typename T2>
void mod_add(T1 &a, T2 b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
struct segment_change {
  int new_w;
  int a;
  segment_change(int _new_w = 0, int _a = 0) : new_w(_new_w), a(_a) {}
};
struct segment {
  long long w_sum;
  int aw_sum;
  segment() : w_sum(0), aw_sum(0) {}
  void apply(int start, int end, const segment_change &change) {
    assert(end - start == 1);
    w_sum = change.new_w;
    aw_sum = (long long)change.a * change.new_w % MOD;
  }
  void join(const segment &other) {
    w_sum += other.w_sum;
    mod_add(aw_sum, other.aw_sum);
  }
  void join(const segment &a, const segment &b) {
    *this = a;
    join(b);
  }
};
struct seg_tree {
  int tree_n;
  vector<segment> tree;
  seg_tree(int n = 0) {
    tree_n = 0;
    if (n > 0) init(n);
  }
  void init(int n) {
    tree_n = 1;
    while (tree_n < n) tree_n *= 2;
    tree.assign(2 * tree_n, segment());
  }
  template <typename T_range_op, typename T_join_op>
  void process_range(int position, int start, int end, int a, int b,
                     T_range_op &&range_op, T_join_op &&join_op) {
    if (a <= start && end <= b) {
      range_op(position, start, end);
      return;
    }
    if (position >= tree_n) return;
    int mid = (start + end) / 2;
    if (a < mid)
      process_range(2 * position, start, mid, a, b, range_op, join_op);
    if (b > mid)
      process_range(2 * position + 1, mid, end, a, b, range_op, join_op);
    join_op(position);
  }
  segment query(int a, int b) {
    segment answer;
    process_range(
        1, 0, tree_n, a, b,
        [&](int position, int, int) { answer.join(tree[position]); },
        [&](int) {});
    return answer;
  }
  void update(int a, int b, const segment_change &change) {
    process_range(
        1, 0, tree_n, a, b,
        [&](int position, int start, int end) {
          tree[position].apply(start, end, change);
        },
        [&](int position) {
          tree[position].join(tree[2 * position], tree[2 * position + 1]);
        });
  }
};
int N, Q;
vector<int> A, W;
seg_tree tree;
int solve(int start, int end) {
  long long total_w = tree.query(start, end).w_sum;
  int low = start, high = end - 1;
  while (low < high) {
    int mid = (low + high) / 2;
    if (2 * tree.query(start, mid + 1).w_sum >= total_w)
      high = mid;
    else
      low = mid + 1;
  }
  segment left = tree.query(start, low);
  segment right = tree.query(low, end);
  long long result = (left.w_sum - right.w_sum) % MOD * (long long)A[low] +
                     right.aw_sum - left.aw_sum;
  return (result % MOD + MOD) % MOD;
}
int main() {
  scanf("%d %d", &N, &Q);
  A.resize(N);
  W.resize(N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    A[i] -= i;
  }
  tree.init(N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &W[i]);
    tree.update(i, i + 1, segment_change(W[i], A[i]));
  }
  for (int q = 0; q < Q; q++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      x = -x - 1;
      tree.update(x, x + 1, segment_change(y, A[x]));
    } else {
      printf("%d\n", solve(x - 1, y));
    }
  }
}