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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 2e5 + 10;
template <class T>
struct range_bit {
  T B1[N];
  T query(int b) {
    T sum = 0;
    for (; b; b -= (b & (-b))) sum += B1[b];
    return sum;
  }
  T range_query(int i, int j) { return query(j) - query(i - 1); }
  void update(int k, T v) {
    for (; k < N; k += (k & (-k))) B1[k] += v;
  }
};
range_bit<long long> bit1, bit2;
range_bit<long double> bit3;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
int a[N], w[N];
long double get(int l, int r, int pos) {
  return bit3.range_query(pos + 1, r) - bit3.range_query(l, pos) +
         a[pos] * (long double)(bit1.range_query(l, pos) -
                                bit1.range_query(pos + 1, r));
}
int get_mod(int l, int r, int pos) {
  long long x1 = bit2.range_query(pos + 1, r) - bit2.range_query(l, pos);
  x1 %= mod;
  x1 += mod;
  x1 %= mod;
  long long x2 = bit1.range_query(l, pos) - bit1.range_query(pos + 1, r);
  x2 %= mod;
  x2 += mod;
  x2 %= mod;
  return add(x1, mul(a[pos], x2));
}
int main() {
  int n, q;
  scanf("%d", &(n));
  scanf("%d", &(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]));
    bit1.update(i, w[i]);
    bit2.update(i, mul(w[i], a[i]));
    bit3.update(i, a[i] * (long long)w[i]);
  }
  while (q--) {
    int x, y;
    scanf("%d", &(x));
    scanf("%d", &(y));
    if (x < 0) {
      x *= -1;
      bit1.update(x, y - w[x]);
      bit2.update(x, mul(y, a[x]) - mul(w[x], a[x]));
      bit3.update(x, (y - w[x]) * (long long)a[x]);
      w[x] = y;
    } else {
      int lo = x, hi = y;
      long long total_sum = bit1.range_query(x, y);
      while (lo < hi) {
        int mid = (lo + hi) >> 1;
        if (2 * bit1.range_query(x, mid) >= total_sum) {
          hi = mid;
        } else
          lo = mid + 1;
      }
      if (x == lo) {
        printf("%d\n", get_mod(x, y, x));
      } else if (get(x, y, lo) > get(x, y, lo - 1)) {
        printf("%d\n", get_mod(x, y, lo - 1));
      } else
        printf("%d\n", get_mod(x, y, lo));
    }
  }
}