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>
class mint {
 public:
  unsigned int v;
  static unsigned int _addu(unsigned int a, unsigned int b) {
    return (a + b) % ((int)1e9 + 7);
  }
  static unsigned int _mulu(unsigned int a, unsigned int b) {
    return (unsigned long long int)a * b % ((int)1e9 + 7);
  }
  mint() {}
  mint(unsigned int _v) : v(_v) {}
  mint(int _v) : v((_v % ((int)1e9 + 7) + ((int)1e9 + 7)) % ((int)1e9 + 7)) {}
  mint(unsigned long long int _v) : v(_v % ((int)1e9 + 7)) {}
  mint(long long int _v)
      : v((_v % ((int)1e9 + 7) + ((int)1e9 + 7)) % ((int)1e9 + 7)) {}
  mint &operator+=(const mint &x) {
    v = _addu(v, x.v);
    return *this;
  }
  mint &operator-=(const mint &x) {
    v = _addu(v, ((int)1e9 + 7) - x.v);
    return *this;
  }
  mint &operator*=(const mint &x) {
    v = _mulu(v, x.v);
    return *this;
  }
  mint operator+(const mint &x) { return mint(_addu(v, x.v)); }
  mint operator-(const mint &x) { return mint(_addu(v, ((int)1e9 + 7) - x.v)); }
  mint operator*(const mint &x) { return mint(_mulu(v, x.v)); }
  operator unsigned int() const { return v; }
  operator int() const { return v; }
  operator unsigned long long int() const { return v; }
  operator long long int() const { return v; }
};
template <class T>
class bit {
 public:
  T a[262144];
  void add(int x, T k) {
    while (x < 262144) {
      a[x] += k;
      x |= x + 1;
    }
  }
  T qry(int x) {
    T ans = 0;
    while (x >= 0) {
      ans += a[x];
      x -= x + 1 & ~x;
    }
    return ans;
  }
  int lbd(T k) {
    if (a[262144 - 1] < k) return 262144;
    T ans = -1;
    int x = 262144 >> 1;
    while (x) {
      if (a[ans + x] < k) k -= a[ans += x];
      x >>= 1;
    }
    return ans + 1;
  }
  int ubd(T k) {
    if (a[262144 - 1] <= k) return 262144;
    T ans = -1;
    int x = 262144 >> 1;
    while (x) {
      if (a[ans + x] <= k) k -= a[ans += x];
      x >>= 1;
    }
    return ans + 1;
  }
};
int a[262144], w[262144];
bit<long long int> pw;
bit<mint> pc;
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) scanf("%d", a + i), a[i] -= i;
  for (int i = 0; i < n; ++i) scanf("%d", w + i);
  for (int i = 0; i < n; ++i) pw.add(i, w[i]);
  for (int i = 0; i < n; ++i) pc.add(i, mint(w[i]) * mint(a[i]));
  while (q--) {
    int ins, ex;
    scanf("%d%d", &ins, &ex);
    if (ins < 0) {
      int id = -ins - 1, d = ex - w[id];
      pw.add(id, d), pc.add(id, mint(d) * mint(a[id])), w[id] += d;
    } else {
      int l = ins - 1, r = ex - 1;
      int x = pw.lbd((pw.qry(l - 1) + pw.qry(r) + 1) / 2);
      0;
      mint tx(a[x]);
      mint cl = (tx * mint(pw.qry(x - 1) - pw.qry(l - 1)) -
                 (pc.qry(x - 1) - pc.qry(l - 1)));
      mint cr = ((pc.qry(r) - pc.qry(x)) - tx * mint(pw.qry(r) - pw.qry(x)));
      printf("%d\n", (int)(cl + cr));
    }
  }
  return 0;
}