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 MXN = 2e5 + 10, MXM = 5010;
long long MOD = 1e9 + 7;
int N, Q;
long long a[MXN], w[MXN];
template <typename T>
struct fenwick {
  T f[MXN] = {};
  void update(int x, T val) {
    for (; x <= N; x += x & -x) (f[x] = f[x] + val);
  }
  T get(int x) {
    T res = 0;
    for (; x; x -= x & -x) {
      (res = res + f[x]);
    }
    return res;
  }
  T get(int x1, int x2) {
    x1 = max(x1 - 1, 0);
    x2 = min(x2, N);
    return get(x2) - get(x1);
  }
  T findmid(int x1, int x2) {
    long long s1, s2, goal, res = 0;
    int x = 0;
    x1 = max(x1 - 1, 0);
    x2 = min(x2, N);
    s1 = get(x1);
    s2 = get(x2);
    goal = (s1 + s2 + 1) >> 1;
    for (int k = 20; k >= 0; --k) {
      if (x + (1 << k) <= N)
        if (res + f[x + (1 << k)] < goal) {
          res += f[x + (1 << k)];
          x += 1 << k;
        }
    }
    return x + 1;
  }
};
template <typename T>
struct fenwickMOD {
  T f[MXN] = {};
  void update(int x, T val) {
    for (; x <= N; x += x & -x) (f[x] = f[x] + val) %= MOD;
  }
  T get(int x) {
    T res = 0;
    for (; x; x -= x & -x) {
      (res = res + f[x]) %= MOD;
    }
    return res;
  }
  T get(int x1, int x2) {
    x1 = max(x1 - 1, 0);
    x2 = min(x2, N);
    return (get(x2) - get(x1) + MOD) % MOD;
  }
};
fenwick<long long> F;
fenwickMOD<long long> F2;
void upd(int x, int y) {
  int d = y - w[x];
  F.update(x, d);
  F2.update(x, (a[x] - x) * d);
  w[x] = y;
}
void init() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= N; i++) {
    int x;
    cin >> x;
    upd(i, x);
  }
}
namespace solve {
void solve() {
  for (int t = 0; t < Q; t++) {
    int x, y;
    cin >> x >> y;
    if (x > 0) {
      long long ans = 1e18;
      int m = F.findmid(x, y);
      {
        long long t =
            (F.get(x, m - 1) - F.get(m + 1, y)) % MOD * (a[m] - m) % MOD;
        long long t2 = F2.get(m + 1, y) - F2.get(x, m - 1);
        ans = min(ans, t + t2);
      }
      cout << (ans % MOD + MOD) % MOD << endl;
    } else {
      x = -x;
      upd(x, y);
    }
  }
}
}  // namespace solve
int main() {
  init();
  solve::solve();
  return 0;
}