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;
using ll = long long;
const int mod = 1e9 + 7, nax = 2e5 + 10;
void plusle(int &a, int b) {
  if ((a += b) >= mod) a -= mod;
}
void minun(int &a, int b) {
  if ((a -= b) < 0) a += mod;
}
int add(int a, int b) { return ((a += b) >= mod) ? a - mod : a; }
int sub(int a, int b) { return ((a -= b) < 0) ? a + mod : a; }
int mul(int a, int b) { return (1LL * a * b) % mod; }
template <class T = ll>
struct fenwick {
  T bit[nax];
  T get(int idx) {
    T res = 0;
    for (; idx > 0; idx -= idx & -idx) res += bit[idx];
    return res;
  }
  void update(int idx, int val) {
    for (; idx < nax; idx += idx & -idx) bit[idx] += val;
  }
  T get(int l, int r) {
    if (l > r) return 0;
    return get(r) - get(l - 1);
  }
};
struct modfenwick {
  int bit[nax];
  modfenwick() { memset(bit, 0, sizeof(bit)); }
  int get(int idx) {
    int res = 0;
    for (; idx > 0; idx -= idx & -idx) plusle(res, bit[idx]);
    return res;
  }
  void update(int idx, int val) {
    for (; idx < nax; idx += idx & -idx) plusle(bit[idx], val);
  }
  int get(int l, int r) {
    if (l > r) return 0;
    return sub(get(r), get(l - 1));
  }
};
int n, q, a[nax], w[nax];
modfenwick fm;
fenwick<ll> f1;
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", w + i);
    f1.update(i, w[i]);
    fm.update(i, mul(a[i], w[i]));
  }
  42;
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      x = -x;
      int res = ((-mul(a[x], w[x])) + mod) % mod;
      fm.update(x, res);
      f1.update(x, -w[x]);
      w[x] = y;
      f1.update(x, y);
      fm.update(x, mul(a[x], y));
    } else {
      int l = x, r = y, ans = x;
      ll sum = f1.get(l, r);
      while (l <= r) {
        int mid = l + r >> 1;
        if (f1.get(x, mid) * 2 >= sum) {
          ans = mid;
          r = mid - 1;
        } else
          l = mid + 1;
      }
      42;
      ll res = 0;
      int c = ans;
      if (x < c) {
        res = (1LL * a[c] * (f1.get(x, c) % mod)) % mod;
        42;
        res -= fm.get(x, c);
        42;
      }
      if (y > c) {
        res -= (1LL * a[c] * (f1.get(c + 1, y) % mod)) % mod;
        42;
        res += fm.get(c + 1, y);
        42;
        res %= mod;
      }
      if (res < 0) res += mod;
      42;
      printf("%lld\n", res);
    }
  }
}