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 N = 200200;
const int Mod = 1e9 + 7;
void add_self(int& x, int y) {
  if ((x += y) >= Mod) x -= Mod;
}
void sub_self(int& x, int y) {
  if ((x -= y) < 0) x += Mod;
}
int add(int x, int y) { return add_self(x, y), x; }
int sub(int x, int y) { return sub_self(x, y), x; }
int mul(int x, int y) { return (long long)x * y % Mod; }
template <typename type>
struct SegmentTreeWithoutLazyPropagation {
 private:
  const type E = 0;

 private:
  static const int Nax = 4 * N;

 private:
  int n;

 private:
  type Tree[Nax];

 public:
  SegmentTreeWithoutLazyPropagation(int n = 0) : n(n) {
    for (int Node = 0; Node < Nax; Node++) Tree[Node] = E;
  }

 private:
  virtual type Unite(type x, type y) = 0;

 private:
  void Merge(int Node) {
    Tree[Node] = Unite(Tree[(Node << 1)], Tree[(Node << 1 | 1)]);
  }

 public:
  void Build(const vector<type>& a) {
    Clear();
    n = a.size();
    Build(a, 1, 0, n - 1);
  }

 private:
  void Build(const vector<type>& a, int Node, int L, int R) {
    Tree[Node] = E;
    if (L == R) return void(Tree[Node] = a[L]);
    Build(a, (Node << 1), L, ((L + R) >> 1));
    Build(a, (Node << 1 | 1), ((L + R) >> 1) + 1, R);
    Merge(Node);
  }

 public:
  void Update(int i, type x) {
    if (n <= 0) return;
    Update(i, x, 1, 0, n - 1);
  }

 private:
  void Update(int i, type x, int Node, int L, int R) {
    if (i < L || R < i) return;
    if (L == R) return void(Tree[Node] = x);
    Update(i, x, (Node << 1), L, ((L + R) >> 1));
    Update(i, x, (Node << 1 | 1), ((L + R) >> 1) + 1, R);
    Merge(Node);
  }

 public:
  type Query(int i, int j) {
    if (n <= 0 || i > j) return E;
    return Query(i, j, 1, 0, n - 1);
  }

 private:
  type Query(int i, int j, int Node, int L, int R) {
    if (j < L || R < i) return E;
    if (i <= L && R <= j) return Tree[Node];
    return Unite(Query(i, j, (Node << 1), L, ((L + R) >> 1)),
                 Query(i, j, (Node << 1 | 1), ((L + R) >> 1) + 1, R));
  }

 public:
  void Clear() {
    for (int Node = 0; Node <= 4 * n; Node++) Tree[Node] = E;
    n = 0;
  }
};
template <typename type>
struct SegmentTreeSum : public SegmentTreeWithoutLazyPropagation<type> {
 private:
  type Unite(type x, type y) { return x + y; };
};
template <typename type>
struct SegmentTreeSumMod : public SegmentTreeWithoutLazyPropagation<type> {
 private:
  type Unite(type x, type y) { return add(x, y); };
};
SegmentTreeSum<long long> MyTreeW;
SegmentTreeSumMod<int> MyTreeWA;
int main() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int& x : a) scanf("%d", &x);
  vector<long long> w(n);
  for (long long& x : w) scanf("%I64d", &x);
  vector<int> WA(n);
  for (int i = 0; i < n; i++) WA[i] = mul(w[i], sub(a[i], i));
  MyTreeW.Build(w);
  MyTreeWA.Build(WA);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      int i = -x - 1, nw = y;
      w[i] = nw;
      MyTreeW.Update(i, nw);
      MyTreeWA.Update(i, mul(nw, sub(a[i], i)));
    } else {
      int L = x - 1, R = y - 1;
      int i = L - 1;
      int Low = L, High = R - 1;
      long long Sum = MyTreeW.Query(L, R);
      while (Low <= High) {
        int Mid = Low + (High - Low) / 2;
        long long S = MyTreeW.Query(L, Mid);
        if (2 * S <= Sum) {
          i = Mid;
          Low = Mid + 1;
        } else {
          High = Mid - 1;
        }
      }
      i++;
      int Ans = 0;
      add_self(Ans, mul(a[i] - i, MyTreeW.Query(L, i - 1) % Mod));
      sub_self(Ans, MyTreeWA.Query(L, i - 1));
      sub_self(Ans, mul(a[i] - i, MyTreeW.Query(i + 1, R) % Mod));
      add_self(Ans, MyTreeWA.Query(i + 1, R));
      printf("%d\n", Ans);
    }
  }
}