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 maxn = 2e5 + 10, mod = 1e9 + 7;
int n, m, k, t, a[maxn], w[maxn], lg[maxn];
long long d[2][maxn];
void add(long long d[], int x, int y, bool isMOD) {
  if (isMOD && y >= mod) y -= mod;
  while (x <= n) {
    d[x] += y;
    if (isMOD && d[x] >= mod) d[x] -= mod;
    x += x & (-x);
  }
}
long long get(long long d[], int x, bool isMOD) {
  long long ret = 0;
  while (x) {
    ret += d[x];
    if (isMOD && d[x] >= mod) d[x] -= mod;
    x -= x & (-x);
  }
  return ret;
}
int main() {
  int i, j;
  int q;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] -= i;
    lg[i] = 1 + lg[i >> 1];
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    add(d[0], i, w[i], false);
    add(d[1], i, 1LL * a[i] * w[i] % mod, true);
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (l < 0) {
      l *= -1;
      add(d[0], l, -w[l] + r, false);
      add(d[1], l, 1LL * a[l] * (r - w[l]) % mod + mod, true);
      w[l] = r;
    } else {
      int pos = 0;
      long long now = 0;
      long long sum = (get(d[0], r, false) + get(d[0], l - 1, false)) / 2;
      for (i = lg[r] - 1; i >= 0; i--) {
        if (pos + (1 << i) <= n && now + d[0][pos + (1 << i)] <= sum)
          pos += (1 << i), now += d[0][pos];
      }
      pos = max(l, min(pos + 1, r));
      long long ret = 0;
      ret += (get(d[0], pos, false) - get(d[0], l - 1, false)) % mod * a[pos] %
             mod;
      ret -= get(d[1], pos, true) - get(d[1], l - 1, true);
      ret += get(d[1], r, true) - get(d[1], pos - 1, true);
      ret -= (get(d[0], r, false) - get(d[0], pos - 1, false)) % mod * a[pos] %
             mod;
      ret = (ret % mod + mod) % mod;
      printf("%lld\n", ret);
    }
  }
  return 0;
}