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 long long mod = 1e9 + 7;
const int mn = 2e5 + 10;
long long bit[mn], bit2[mn], sa[mn];
void upd(int a, long long b) {
  for (; a < mn; a += a & -a) bit[a] += b;
}
void upd2(int a, long long b) {
  for (; a < mn; a += a & -a) bit2[a] += b, bit2[a] %= mod;
}
long long qu(int a) {
  long long b = 0;
  for (; a; a -= a & -a) b += bit[a];
  return b;
}
long long qu2(int a) {
  long long b = 0;
  for (; a; a -= a & -a) b += bit2[a], b %= mod;
  return b;
}
long long a[mn], w[mn];
int sch(long long x) {
  int cur = 0;
  long long sum = 0;
  for (int i = 1 << 17; i; i >>= 1) {
    if (cur + i < mn && sum + bit[cur + i] <= x) sum += bit[cur += i];
  }
  return cur + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i], a[i] -= i;
  for (int i = 1; i <= n; i++)
    cin >> w[i], upd(i, w[i]), upd2(i, w[i] * a[i] % mod);
  while (q--) {
    long long l, r;
    cin >> l >> r;
    if (l < 0) {
      l = -l;
      upd(l, r - w[l]);
      upd2(l, (r - w[l]) * a[l] % mod);
      w[l] = r;
    } else {
      long long ls = qu(l - 1), rs = qu(r);
      long long sum = rs - ls;
      int c = sch(sum / 2 + ls);
      long long cs = qu(c);
      long long ans = (qu2(r) - qu2(c) - (rs - cs) % mod * a[c] % mod) -
                      (qu2(c) - qu2(l - 1) - (cs - ls) % mod * a[c] % mod);
      ans %= mod;
      if (ans < 0) ans += mod;
      printf("%lld\n", ans);
    }
  }
}