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 N = 210005, p = 1000000007;
int read() {
  int f = 1, g = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) g = g * 10 + ch - '0';
  return f * g;
}
int n, q;
long long a[N], w[N];
struct node {
  int l, r;
  long long s, sl, sr;
};
node operator+(node x, node y) {
  node z;
  z.l = x.l;
  z.r = y.r;
  z.s = x.s + y.s;
  z.sl = (x.sl + y.sl + y.s % p * (a[y.l] - a[x.l] - x.r + x.l - 1) % p) % p;
  z.sr = (x.sr + y.sr + x.s % p * (a[y.r] - y.r + y.l - 1 - a[x.r]) % p) % p;
  return z;
}
struct segment_tree {
  node a[N << 2];
  void build(int x, int fl, int fr) {
    if (fl == fr) {
      a[x] = (node){fl, fr, w[fl], 0, 0};
      return;
    }
    int mid = (fl + fr) >> 1;
    build(x * 2, fl, mid);
    build(x * 2 + 1, mid + 1, fr);
    a[x] = a[x * 2] + a[x * 2 + 1];
  }
  void modify(int x, int k, long long w) {
    if (a[x].l == a[x].r) {
      a[x].s = w;
      return;
    }
    int mid = (a[x].l + a[x].r) >> 1;
    modify(x * 2 + (k > mid), k, w);
    a[x] = a[x * 2] + a[x * 2 + 1];
  }
  long long getsum(int x, int k) {
    if (a[x].r <= k) return a[x].s;
    int mid = (a[x].l + a[x].r) >> 1;
    long long s = getsum(x * 2, k);
    if (mid < k) s += getsum(x * 2 + 1, k);
    return s;
  }
  int find(int x, long long k) {
    if (a[x].l == a[x].r) return a[x].l;
    int mid = (a[x].l + a[x].r) >> 1;
    if (k <= a[x * 2].s)
      return find(x * 2, k);
    else
      return find(x * 2 + 1, k - a[x * 2].s);
  }
  int lower_bound(long long k) { return find(1, k); }
  node query(int x, int l, int r) {
    if ((l <= a[x].l) && (a[x].r <= r)) return a[x];
    int mid = (a[x].l + a[x].r) >> 1;
    if (r <= mid)
      return query(x * 2, l, r);
    else if (mid < l)
      return query(x * 2 + 1, l, r);
    else
      return query(x * 2, l, r) + query(x * 2 + 1, l, r);
  }
  long long calc(int l, int r) {
    int t = lower_bound((getsum(1, l - 1) + getsum(1, r) + 1) >> 1);
    return (query(1, l, t).sr + query(1, t, r).sl) % p;
  }
} f;
int main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) w[i] = read();
  a[0] = 0;
  a[n + 1] = 0x3f3f3f3f;
  f.build(1, 0, n + 1);
  while (q--) {
    int x = read(), y = read();
    if (x < 0)
      f.modify(1, -x, y);
    else
      printf("%d\n", f.calc(x, y));
  }
  return 0;
}