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;
using ll = long long;
template <typename T>
class fenwick_tree {
  const int n;
  vector<T> data;

 public:
  fenwick_tree(int n_) : n(n_), data(n + 1) {}
  T find(int p) const {
    T res = 0;
    while (p > 0) {
      res += data[p];
      p -= p & -p;
    }
    return res;
  }
  T find(int l, int r) const { return find(r) - find(l); }
  void add(int p, T val) {
    ++p;
    while (p <= n) {
      data[p] += val;
      p += p & -p;
    }
  }
  void update(int p, T val) { add(p, val - find(p, p + 1)); }
};
template <int MOD>
class mod_int {
  unsigned x;

 public:
  mod_int() : x(0) {}
  mod_int(int sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  mod_int(long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  mod_int &operator+=(mod_int that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator-=(mod_int that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator*=(mod_int that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  mod_int &operator/=(mod_int that) { return *this *= that.inverse(); }
  mod_int operator+(mod_int that) const { return mod_int(*this) += that; }
  mod_int operator-(mod_int that) const { return mod_int(*this) -= that; }
  mod_int operator*(mod_int that) const { return mod_int(*this) *= that; }
  mod_int operator/(mod_int that) const { return mod_int(*this) /= that; }
  bool operator==(const mod_int &that) const { return x == that.x; }
  mod_int inverse() const {
    long long a = x, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return mod_int(u);
  }
};
const int mod = 1e9 + 7;
using mint = mod_int<mod>;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  fenwick_tree<ll> co(n);
  fenwick_tree<mint> all(n);
  for (int i = 0; i < n; i++) {
    int w;
    cin >> w;
    co.add(i, w);
    all.add(i, mint(1) * w * a[i]);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      int id = -x - 1;
      co.update(id, y);
      all.update(id, mint(1) * y * a[id]);
    } else {
      --x;
      ll sum = co.find(x, y);
      int lb = x, ub = y;
      while (ub - lb > 1) {
        int m = (lb + ub) / 2;
        (co.find(x, m) * 2 <= sum ? lb : ub) = m;
      }
      mint pos = a[lb];
      ll lc = co.find(x, lb), rc = sum - lc;
      mint res = (pos * lc - all.find(x, lb)) + (all.find(lb, y) - pos * rc);
      printf("%d\n", res.get());
    }
  }
  return 0;
}