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 T>
struct BIT {
  int n;
  vector<T> t;
  BIT(int n) : n(n), t(n + 1, 0) {}
  void modify(int r, T v) {
    for (int i = r; i <= n; i += i & -i) t[i] += v;
  }
  T sum(int r) {
    T s = 0;
    for (int i = r; i >= 1; i -= i & -i) s += t[i];
    return s;
  }
  T sum(int l, int r) { return sum(r) - sum(l - 1); }
  int lower_bound(T v) {
    T sum = 0, pos = 0, k = 31 - __builtin_clz(n);
    for (int i = k; i >= 0; i--) {
      if (pos + (1 << i) <= n && sum + t[pos + (1 << i)] < v) {
        pos += 1 << i;
        sum += t[pos];
      }
    }
    return pos + 1;
  }
};
template <int mod = 1000000007>
struct modint {
  int x;
  modint(long long y = 0) : x((y % mod + mod) % mod) {}
  friend modint operator^(modint a, int64_t b) {
    modint r = 1;
    for (; b; b >>= 1, a *= a)
      if (b & 1) r *= a;
    return r;
  }
  friend modint operator-(modint a) { return modint(0) - a; }
  friend modint operator!(modint a) { return a ^ (mod - 2); }
  modint& operator/=(modint const& b) { return *this *= !b; }
  friend modint operator+(modint a, modint const& b) { return a += b; }
  friend modint operator-(modint a, modint const& b) { return a -= b; }
  friend modint operator*(modint a, modint const& b) { return a *= b; }
  friend modint operator/(modint a, modint const& b) { return a /= b; }
  friend bool operator!=(const modint& a, const modint b) { return a.x != b.x; }
  friend bool operator==(const modint& a, const modint b) { return a.x == b.x; }
  modint& operator*=(modint const& b) {
    x = 1LL * x * b.x % mod;
    return *this;
  }
  friend ostream& operator<<(ostream& os, modint const& a) { return os << a.x; }
  modint& operator+=(modint const& b) {
    x += b.x;
    x = (x >= mod) ? x - mod : x;
    return *this;
  }
  modint& operator-=(modint const& b) {
    x = x >= b.x ? x - b.x : x - b.x + mod;
    return *this;
  }
};
using mint = modint<>;
using LL = long long;
const int N = 2e5 + 10;
int a[N], w[N];
void update(int i, int x, BIT<LL>& bit, BIT<mint>& t) {
  bit.modify(i, x - w[i]);
  t.modify(i, mint(x - w[i]) * mint(a[i] - i));
  w[i] = x;
}
mint cost(int l, int r, int m, BIT<LL>& W, BIT<mint>& WX) {
  return mint(a[m] - m) * mint(W.sum(l, m)) - WX.sum(l, m) + WX.sum(m, r) -
         mint(a[m] - m) * mint(W.sum(m, r));
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  BIT<LL> bit(n);
  BIT<mint> t(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    update(i, x, bit, t);
  }
  for (int i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    if (x < 0)
      update(-x, y, bit, t);
    else {
      int idx = bit.lower_bound(bit.sum(x - 1) + bit.sum(x, y) / 2 + 1);
      cout << cost(x, y, idx, bit, t) << endl;
    }
  }
}