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;
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
const int N = 2e5 + 5;
const int mod = 1000000007;
int n, q;
long long a[N], w[N], c1[N], c2[N];
void mdf1(int k, long long v) {
  while (k <= n) c1[k] += v, k += k & -k;
}
long long qry1(int k) {
  long long res = 0;
  while (k) res += c1[k], k ^= k & -k;
  return res;
}
void mdf2(int k, long long v) {
  while (k <= n) c2[k] = (c2[k] + v) % mod, k += k & -k;
}
long long qry2(int k) {
  long long res = 0;
  while (k) res = (res + c2[k]) % mod, k ^= k & -k;
  return res;
}
long long cal(int l, int r, int p, int op) {
  long long s = (qry1(r) - qry1(l - 1)) % mod,
            ss = (qry2(r) - qry2(l - 1) + mod) % mod;
  if (!op)
    return (ss + s * (p - r + mod)) % mod;
  else
    return (s * (l - p + mod) - ss + mod) % mod;
}
int main() {
  n = gi();
  q = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i) w[i] = gi();
  for (int i = 1; i <= n; ++i)
    mdf1(i, w[i]), mdf2(i, w[i] * (i - a[i] + mod) % mod);
  while (q--) {
    int x = gi(), y = gi();
    if (x < 0)
      x = -x, mdf1(x, y - w[x]),
      mdf2(x, (y - w[x] + mod) * (x - a[x] + mod) % mod), w[x] = y;
    else {
      int l = x, r = y, res = x;
      long long sum1 = qry1(x - 1), sum2 = qry1(y) - sum1;
      while (l <= r) {
        int mid = l + r >> 1;
        if ((qry1(mid) - sum1) * 2 >= sum2)
          res = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      long long ans = cal(x, res, a[res], 0) + cal(res + 1, y, a[res] + 1, 1);
      ans %= mod;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}