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 min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
int read() {
  int x = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  return x * w;
}
const int N = 2e5 + 10, Mod = 1e9 + 7;
int n, q, a[N], w[N], c2[N];
long long c1[N];
void add1(int x, long long val) {
  for (; x <= n; x += ((x) & (-x))) c1[x] += val;
}
void add2(int x, int val) {
  for (; x <= n; x += ((x) & (-x))) (c2[x] += val) %= Mod;
}
long long query1(int x) {
  long long res = 0;
  for (; x; x -= ((x) & (-x))) res += c1[x];
  return res;
}
int query2(int x) {
  int res = 0;
  for (; x; x -= ((x) & (-x))) (res += c2[x]) %= Mod;
  return res;
}
int main() {
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  n = read();
  q = read();
  for (int i = 1; i <= n; i++) a[i] = read() - i + n;
  for (int i = 1; i <= n; i++) {
    w[i] = read();
    add1(i, w[i]);
    add2(i, 1ll * w[i] * a[i] % Mod);
  }
  while (q--) {
    int x = read(), y = read();
    if (x < 0) {
      x = -x;
      add1(x, -w[x]);
      add2(x, (-1ll * w[x] * a[x] % Mod + Mod) % Mod);
      w[x] = y;
      add1(x, w[x]);
      add2(x, 1ll * w[x] * a[x] % Mod);
    } else {
      long long s1 = query1(x - 1), s2 = query1(y) - s1;
      int l = x, r = y, pos = y;
      while (l <= r) {
        int mid = (l + r) >> 1;
        long long now = query1(mid) - s1;
        if (now * 2 >= s2)
          r = mid - 1, pos = mid;
        else
          l = mid + 1;
      }
      l = x;
      r = y;
      long long t1 = query2(pos) - query2(l - 1);
      long long t2 = -1ll * a[pos] * ((query1(pos) - s1) % Mod) % Mod;
      long long t3 = query2(r) - query2(pos);
      long long t4 = -1ll * a[pos] * ((query1(r) - query1(pos)) % Mod) % Mod;
      long long ans = (t3 + t4 - t1 - t2) % Mod;
      ans = (ans + Mod) % Mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}