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 <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
  bool done_something = false;
  stringstream res;
  res << "[";
  for (; begin_iter != end_iter and counter; ++begin_iter) {
    done_something = true;
    counter--;
    res << *begin_iter << ", ";
  }
  string str = res.str();
  if (done_something) {
    str.pop_back();
    str.pop_back();
  }
  str += "]";
  return str;
}
vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
  vector<int> ord(size);
  for (int i = 0; i < size; i++) ord[i] = i;
  sort(ord.begin(), ord.end(), compare);
  return ord;
}
template <typename T>
bool MinPlace(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename Container>
int SZ(const Container& S) {
  return S.size();
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T>& p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
using namespace std;
template <typename Value>
class Fenwick {
 public:
  Fenwick(int N) : size(N), max_pow2(log2(N)), container(N, 0) {}
  Fenwick(const vector<Value>& initial_values)
      : Fenwick(initial_values.size()) {
    for (int p = 0; p < size; p++) {
      int q = p | (p + 1);
      container[p] += initial_values[p];
      if (q < size) container[q] += container[p];
    }
  }
  void Add(int p, Value x) {
    for (; p < size; p |= (p + 1)) container[p] += x;
  }
  Value GetRange(int l, int r) {
    if (l > 0) return GetRange(0, r) - GetRange(0, l - 1);
    Value res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) res += container[r];
    return res;
  }

 private:
  int size;
  int max_pow2;
  vector<Value> container;
};
const long long mod = 1e9 + 7;
template <typename Value>
class FenwickMod {
 public:
  FenwickMod(int N) : size(N), max_pow2(log2(N)), container(N, 0) {}
  FenwickMod(const vector<Value>& initial_values)
      : FenwickMod(initial_values.size()) {
    for (int p = 0; p < size; p++) {
      int q = p | (p + 1);
      container[p] += initial_values[p];
      container[p] %= mod;
      if (q < size) {
        container[q] += container[p];
        container[q] %= mod;
      }
    }
  }
  void Add(int p, Value x) {
    for (; p < size; p |= (p + 1)) {
      container[p] += x;
      container[p] %= mod;
    }
  }
  Value GetRange(int l, int r) {
    if (l > 0) return (GetRange(0, r) - GetRange(0, l - 1)) % mod;
    Value res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) res += container[r];
    res %= mod;
    return res;
  }

 private:
  int size;
  int max_pow2;
  vector<Value> container;
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, Q;
  cin >> N >> Q;
  vector<long long> A(N + 10, 0), W(N + 10, 0), AW(N + 10, 0);
  for (int i = 1; i <= N; i++) cin >> A[i], A[i] -= i;
  for (int i = 1; i <= N; i++) cin >> W[i];
  for (int i = 1; i <= N; i++) AW[i] = A[i] * W[i] % mod;
  Fenwick<long long> fenw(W);
  FenwickMod<long long> fenaw(AW);
  for (int q = 0; q < Q; q++) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      fenw.Add(x, y - W[x]);
      fenaw.Add(x, (A[x] * y - A[x] * W[x]) % mod);
      W[x] = y;
    } else {
      int l = x, r = y;
      while (l < r) {
        int m = (l + r) / 2;
        if (fenw.GetRange(x, m) < fenw.GetRange(m + 1, y))
          l = m + 1;
        else
          r = m;
      }
      int m = l;
      l = x, r = y;
      long long res1 = A[m] * (fenw.GetRange(l, m) % mod) % mod;
      long long res2 = fenaw.GetRange(l, m);
      long long res3 = A[m] * (fenw.GetRange(m, r) % mod) % mod;
      long long res4 = fenaw.GetRange(m, r);
      long long res = res1 - res2 - res3 + res4;
      res %= mod;
      if (res < 0) res += mod;
      cout << res << "\n";
    }
  }
}