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 i64 = long long;
template <class T = long long int>
class Bit {
 private:
  unsigned len;
  T init;
  std::vector<T> arr;

 public:
  Bit(unsigned length, T initialValue = 0)
      : len(length), init(initialValue), arr(len + 1, init) {}
  void update(unsigned a, T newval) {
    for (unsigned x = a; x <= len; x += x & -x) arr[x] += newval;
  }
  T query(unsigned a) const {
    T ret = init;
    for (unsigned x = a; x > 0; x -= x & -x) ret += arr[x];
    return ret;
  }
};
int main() {
  constexpr i64 mod = 1000000007;
  int n, q;
  scanf("%d%d", &n, &q);
  std::vector<i64> a(n), w(n);
  for (auto &e : a) scanf("%lld", &e);
  for (int i = 0; i < n; i++) a[i] -= i;
  for (auto &e : w) scanf("%lld", &e);
  Bit<> ss(n + 10), pr(n + 10);
  for (int i = 0; i < n; i++) {
    ss.update(i + 1, w[i]);
    pr.update(i + 1, w[i] * a[i] % mod);
  }
  while (q--) {
    i64 x, y;
    scanf("%lld%lld", &x, &y);
    if (x < 0) {
      const int i = -x - 1;
      ss.update(i + 1, y - w[i]);
      pr.update(i + 1, y * a[i] % mod - w[i] * a[i] % mod);
      w[i] = y;
    } else {
      if (x == y) {
        printf("0\n");
        continue;
      }
      int l = x - 1, r = y;
      const i64 lb = ss.query(x - 1), sum = ss.query(y) - lb;
      while (r - l > 1) {
        const int k = (l + r) / 2;
        const i64 s = ss.query(k) - lb;
        if (2 * s <= sum)
          l = k;
        else
          r = k;
      }
      const i64 lss = (ss.query(l) % mod + mod - lb % mod) % mod * a[l] % mod,
                rss = (ss.query(y) % mod + mod - ss.query(l) % mod) % mod *
                      a[l] % mod,
                lpr = (pr.query(l) % mod + mod - pr.query(x - 1) % mod) % mod,
                rpr = (pr.query(y) % mod + mod - pr.query(l) % mod) % mod;
      const i64 ret = (lss - lpr + 2 * mod + rpr - rss) % mod;
      printf("%lld\n", ret);
    }
  }
  return 0;
}