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>
const int mod = 1000000007;
int n, a[200001], x, y, T[200001], q, l, r, mid, ans, Ans, w[200001],
    t1[200001], t2[200001];
long long t[200001];
int plus(int x, int y) { return (x + y >= mod) ? (x + y - mod) : (x + y); }
void add(int x, long long val) {
  for (; x <= n; x += (x & -x)) t[x] += val;
}
long long query(int x) {
  long long ans = 0;
  for (; x; x -= (x & -x)) ans += t[x];
  return ans;
}
void Add(int x, int val) {
  for (; x <= n; x += (x & -x)) T[x] = plus(T[x], val);
}
int Query(int x) {
  int ans = 0;
  for (; x; x -= (x & -x)) ans = plus(ans, T[x]);
  return ans;
}
long long calc(int l, int r) {
  if (l > r) return 0;
  return query(r) - query(l - 1);
}
int Calc(int l, int r) {
  if (l > r) return 0;
  return (Query(r) - Query(l - 1) + mod) % mod;
}
void add1(int x, int val) {
  for (; x <= n; x += (x & -x)) t1[x] = plus(t1[x], val);
}
int query1(int x) {
  int ans = 0;
  for (; x; x -= (x & -x)) ans = plus(t1[x], ans);
  return ans;
}
void add2(int x, int val) {
  for (; x <= n; x += (x & -x)) t2[x] = plus(t2[x], val);
}
int query2(int x) {
  int ans = 0;
  for (; x; x -= (x & -x)) ans = plus(t2[x], ans);
  return ans;
}
int calc1(int l, int r) {
  if (l > r) return 0;
  return (query1(r) - query1(l - 1) + mod) % mod;
}
int calc2(int l, int r) {
  if (l > r) return 0;
  return (query2(r) - query2(l - 1) + mod) % mod;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(t, 0, sizeof(t));
  memset(T, 0, sizeof(T));
  memset(t1, 0, sizeof(t1));
  memset(t2, 0, sizeof(t2));
  for (int i = 1; i <= n; i++)
    scanf("%d", &x), add(i, x), Add(i, 1ll * x * a[i] % mod),
        w[i] = x, add1(i, 1ll * i * x % mod),
        add2(i, 1ll * (n - i + 1) * x % mod);
  while (q--) {
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      add(x, y - w[x]);
      Add(x, 1ll * (y - w[x] + mod) * a[x] % mod);
      add1(x, 1ll * (y - w[x] + mod) * x % mod);
      add2(x, 1ll * (y - w[x] + mod) * (n - x + 1) % mod);
      w[x] = y;
    } else {
      l = x;
      r = y;
      ans = y + 1;
      while (l <= r) {
        mid = (l + r) >> 1;
        if (calc(x, mid - 1) >= calc(mid, y))
          ans = mid, r = mid - 1;
        else
          l = mid + 1;
      }
      ans--;
      Ans = (calc(x, ans - 1) % mod * a[ans] - Calc(x, ans - 1)) % mod;
      Ans = (-calc(ans + 1, y) % mod * a[ans] + Ans + Calc(ans + 1, y)) % mod;
      Ans = (Ans - calc2(x, ans - 1) + calc(x, ans - 1) % mod * (n - ans + 1)) %
            mod;
      Ans = (Ans - calc1(ans + 1, y) + calc(ans + 1, y) % mod * ans) % mod;
      printf("%d\n", (Ans + mod) % mod);
    }
  }
  return 0;
}