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;
template <class T>
class SegmentTree {
 private:
  vector<T> tree;
  vector<T> data;
  int N;
  inline T OP(T x, T y) { return x + y; }
  T count_sum(int tv, int tl, int tr, int l, int r) {
    if (tl >= tr) return T();
    l = max(l, tl);
    r = min(r, tr);
    if (l >= r) return T();
    if (l == tl && r == tr) return tree[tv];
    int tm = (tl + tr) / 2;
    return OP(count_sum(2 * tv, tl, tm, l, r),
              count_sum(2 * tv + 1, tm, tr, l, r));
  }
  void update_value(int tv, int tl, int tr, int idx, T new_val) {
    if (tl >= tr) return;
    if (tl == idx && tr == idx + 1) {
      tree[tv] = new_val;
      return;
    }
    int tm = (tl + tr) / 2;
    if (idx < tm)
      update_value(2 * tv, tl, tm, idx, new_val);
    else
      update_value(2 * tv + 1, tm, tr, idx, new_val);
    tree[tv] = OP(tree[2 * tv], tree[2 * tv + 1]);
  }
  T build(int tv, int tl, int tr) {
    if (tl >= tr) return T();
    if (tl + 1 == tr) return tree[tv] = data[tl];
    int tm = (tl + tr) / 2;
    return tree[tv] = OP(build(2 * tv, tl, tm), build(2 * tv + 1, tm, tr));
  }

 public:
  SegmentTree() : SegmentTree(vector<T>()) {}
  SegmentTree(int n) : SegmentTree(vector<T>(n)) {}
  SegmentTree(vector<T> data_) {
    data = data_;
    N = data.size();
    tree = vector<T>(4 * N);
    if (N > 0) build(1, 0, N);
  }
  void set(int idx, T new_value) {
    data[idx] = new_value;
    update_value(1, 0, N, idx, new_value);
  }
  T get(int idx) { return data[idx]; }
  T sum(int l, int r) { return count_sum(1, 0, N, max(0, l), min(N, r)); }
};
vector<int64_t> a, b;
SegmentTree<int64_t> w, wb;
int64_t find_ans(int l, int r) {
  int zl = l, zr = r, zm;
  while (zl + 1 != zr) {
    int zm = (zl + zr) / 2;
    int64_t SL = w.sum(l, zm);
    int64_t SR = w.sum(zm, r);
    if (SL - SR <= 0)
      zl = zm;
    else
      zr = zm;
  }
  int z = zl;
  int64_t s1 = (b[z] * (w.sum(l, z) - w.sum(z, r))) % 1000000007;
  int64_t s2 = (wb.sum(l, z) - wb.sum(z, r)) % 1000000007;
  return ((s1 - s2) % 1000000007 + 1000000007) % 1000000007;
}
vector<int64_t> read_vector(int n) {
  vector<int64_t> ans(n);
  for (int i = 0; i < n; i++) cin >> ans[i];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  a = read_vector(n);
  w = SegmentTree<int64_t>(read_vector(n));
  b = vector<int64_t>(n);
  wb = SegmentTree<int64_t>(n);
  for (int i = 0; i < n; i++) {
    b[i] = a[i] - i;
    wb.set(i, (w.get(i) * b[i]) % 1000000007);
  }
  for (int i = 0; i < q; i++) {
    int64_t x, y;
    cin >> x >> y;
    if (x < 0) {
      int idx = (-x) - 1;
      w.set(idx, y);
      wb.set(idx, (y * b[idx]) % 1000000007);
    } else {
      cout << find_ans(x - 1, y) << "\n";
    }
  }
  return 0;
}