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 ul = std::uint32_t;
using ull = std::uint64_t;
using li = std::int32_t;
using ll = std::int64_t;
using vull = std::vector<ull>;
using vul = std::vector<ul>;
const ul sz = 1 << 18;
const ul base = 1e9 + 7;
class scaler {
 public:
  ul v = 0;
  scaler() = default;
  scaler(ul x) : v(x) {}
};
scaler operator+(scaler a, scaler b) {
  return scaler(a.v + b.v < base ? a.v + b.v : a.v + b.v - base);
}
scaler operator-(scaler a, scaler b) {
  return scaler(a.v < b.v ? a.v + base - b.v : a.v - b.v);
}
scaler operator*(scaler a, scaler b) { return scaler(ull(a.v) * b.v % base); }
template <typename T>
void change(ul pos, const T& v, std::vector<T>& tree) {
  for (tree[pos |= sz] = v, pos >>= 1; pos; pos >>= 1) {
    tree[pos] = tree[pos << 1] + tree[pos << 1 | 1];
  }
}
template <typename T>
T query(ul l, ul r, const std::vector<T>& tree) {
  T ret = 0;
  for (l = l - 1 | sz, r = r + 1 | sz; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) {
      ret = ret + tree[l ^ 1];
    }
    if (r & 1) {
      ret = ret + tree[r ^ 1];
    }
  }
  return ret;
}
std::vector<scaler> atree(sz << 1, ul(0));
vull tree(sz << 1, ul(0));
ul n, q;
vul a(2e5 + 1);
scaler compute(ul l, ul r, ul cut) {
  return scaler(query(l, cut - 1, tree) % base) * scaler(a[cut]) -
         query(l, cut - 1, atree) + query(cut + 1, r, atree) -
         scaler(query(cut + 1, r, tree) % base) * scaler(a[cut]);
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cin >> n >> q;
  for (ul i = 1; i <= n; ++i) {
    std::cin >> a[i];
    a[i] -= i;
  }
  for (ul i = 1; i <= n; ++i) {
    ul w;
    std::cin >> w;
    change(i, ull(w), tree);
    change(i, scaler(a[i]) * scaler(w), atree);
  }
  for (ul i = 0; i != q; ++i) {
    li x, y;
    std::cin >> x >> y;
    if (x < 0) {
      ul id = -x;
      ul w = y;
      change(id, ull(w), tree);
      change(id, scaler(a[id]) * scaler(w), atree);
    } else {
      ul l = x, r = y;
      ull temp = query(l, r, tree);
      ul lans = l, rans = r + 1;
      while (lans + 1 != rans) {
        ul mid = lans + rans >> 1;
        if ((query(mid, r, tree) << 1) < temp) {
          rans = mid;
        } else {
          lans = mid;
        }
      }
      std::cout << compute(l, r, lans).v << std::endl;
    }
  }
  return 0;
}