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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
inline unsigned int R() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 444444, mo = 1e9 + 7;
int n, q, a[N], w[N];
long long v1[N], v2[N], v3[N];
void add(int p, long long v) {
  v3[0] += v;
  while (p <= n) v3[p] += v, p += p & -p;
}
long long query(int p) {
  long long res = 0;
  while (p) res += v3[p], p -= p & -p;
  return res;
}
inline void add(long long *a, int p, long long v) {
  a[0] = (a[0] + v) % mo;
  while (p <= n) a[p] = (a[p] + v) % mo, p += p & -p;
}
inline long long query(long long *a, int p) {
  long long res = 0;
  while (p) res += a[p], p -= p & -p;
  return res % mo;
}
long long getval(long long *a, int l, int r) {
  l = max(l, 1);
  r = min(r, n);
  if (l > r) return 0;
  if (l == 1) return query(a, r);
  return (a[0] - query(a, l - 1) + mo) % mo;
}
long long getans(int x, int p) {
  return (query(v1, x) + 1LL * query(v2, x) * (a[p] - p)) % mo;
}
int main() {
  read(n);
  read(q);
  for (register int i = 1; i <= n; i++) read(a[i]);
  for (register int i = 1; i <= n; i++)
    read(w[i]), add(v1, i, 1LL * (i - a[i]) * w[i] % mo), add(v2, i, w[i]),
        add(i, w[i]);
  while (q--) {
    int u, v;
    read(u);
    read(v);
    if (u < 0) {
      u = -u;
      add(v1, u, 1LL * (u - a[u]) * (v - w[u] + mo) % mo);
      add(v2, u, v - w[u]);
      add(u, v - w[u]);
      w[u] = v;
    } else {
      int l = u, r = v, p = l, L = l, R = r;
      while (L <= R) {
        int mid = (L + R) >> 1;
        long long x = query(mid) - query(l - 1) - (query(r) - query(mid));
        if (x == 0) {
          p = mid;
          break;
        }
        if (x < 0) p = mid;
        if (x > 0)
          R = mid - 1;
        else
          L = mid + 1;
      }
      if (p + 1 <= r && query(p) - query(l - 1) < query(r) - query(p)) p++;
      long long res =
          getans(p - 1, p) - getans(l - 1, p) - (getans(r, p) - getans(p, p));
      printf("%lld\n", (res % mo + mo) % mo);
    }
  }
  return 0;
}