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]. 
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int MOD = 1e9 + 7;
struct mint {
  int x;
  mint(long long arg) {
    x = arg % MOD;
    if (x < 0) x += MOD;
  }
  mint() : x(0) {}
  void operator+=(const mint &rhs) {
    x += rhs.x;
    if (x >= MOD) x -= MOD;
  }
  void operator-=(const mint &rhs) {
    x -= rhs.x;
    if (x < 0) x += MOD;
  }
  mint operator+(const mint &rhs) const {
    mint tmp = *this;
    tmp += rhs;
    return tmp;
  }
  mint operator-(const mint &rhs) const {
    mint tmp = *this;
    tmp -= rhs;
    return tmp;
  }
  mint operator*(const mint &rhs) const { return (long long)x * rhs.x % MOD; }
};
template <typename T>
struct fenwick {
  T f[MAXN];
  fenwick() {
    for (int i = 0; i < MAXN; i++) f[i] = 0;
  }
  void update(int x, T val) {
    for (; x < MAXN; x += x & -x) f[x] += val;
  }
  T get(int x) {
    T res = 0;
    for (; x; x -= x & -x) res += f[x];
    return res;
  }
};
int N, Q;
int a[MAXN], w[MAXN];
fenwick<long long> sum;
fenwick<mint> weight;
void load() {
  scanf("%d%d", &N, &Q);
  for (int i = 1; i <= N; i++) scanf("%d", a + i);
  for (int i = 1; i <= N; i++) scanf("%d", w + i);
}
mint query(int l, int r) {
  int lo = l, hi = r + 1;
  long long pref = sum.get(l - 1);
  long long tot = sum.get(r) - pref;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    if (2 * (sum.get(mid) - pref) >= tot)
      hi = mid;
    else
      lo = mid + 1;
  }
  return mint(a[lo]) * mint(2 * sum.get(lo) - sum.get(l - 1) - sum.get(r)) +
         weight.get(l - 1) + weight.get(r) - mint(2) * weight.get(lo);
}
void solve() {
  for (int i = 1; i <= N; i++) {
    a[i] -= i;
    sum.update(i, w[i]);
    weight.update(i, mint(a[i]) * mint(w[i]));
  }
  while (Q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      sum.update(x, y - w[x]);
      weight.update(x, mint(a[x]) * mint(y - w[x]));
      w[x] = y;
    } else
      printf("%d\n", query(x, y).x);
  }
}
int main() {
  load();
  solve();
  return 0;
}
