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 T>
inline T read(T &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int MOD = 1e9 + 7;
int N, Q;
int X[200005], W[200005];
struct VALUE {
  long long sw;
  int swi, swx;
  VALUE operator+(const VALUE &ot) const {
    VALUE ret;
    ret.sw = sw + ot.sw;
    ret.swi = (swi + ot.swi) % MOD;
    ret.swx = (swx + ot.swx) % MOD;
    return ret;
  }
} zero = {0, 0, 0};
struct NODE {
  NODE() : left(0), right(0) {}
  VALUE v;
  void renew() { v = left->v + right->v; }
  NODE *left, *right;
} * root;
NODE *make_tree(int s, int e) {
  NODE *ret = new NODE();
  if (s == e) {
    ret->v = {
        W[s],
        (int)((long long)W[s] * s % MOD),
        (int)((long long)W[s] * X[s] % MOD),
    };
  } else {
    int m = s + e >> 1;
    ret->left = make_tree(s, m);
    ret->right = make_tree(m + 1, e);
    ret->renew();
  }
  return ret;
}
VALUE get(NODE *now, int s, int e, int l, int r) {
  if (e < l || s > r) return zero;
  if (l <= s && e <= r) return now->v;
  int m = s + e >> 1;
  return get(now->left, s, m, l, r) + get(now->right, m + 1, e, l, r);
}
void write(NODE *now, int s, int e, int n, int v) {
  if (e < n || s > n) return;
  if (s == e) {
    now->v = {
        v,
        (int)((long long)v * s % MOD),
        (int)((long long)v * X[s] % MOD),
    };
    return;
  }
  int m = s + e >> 1;
  write(now->left, s, m, n, v);
  write(now->right, m + 1, e, n, v);
  now->renew();
}
int get_mid(NODE *now, int s, int e, int l, int r, long long &sw) {
  if (e < l || s > r || l > r) return 0;
  if (l <= s && e <= r && sw > now->v.sw) {
    sw -= now->v.sw;
    return 0;
  }
  if (s == e) return s;
  int m = s + e >> 1;
  if (int tmp = get_mid(now->left, s, m, l, r, sw)) return tmp;
  return get_mid(now->right, m + 1, e, l, r, sw);
}
int main() {
  read(N, Q);
  for (int i = 1; i <= N; i++) read(X[i]);
  for (int i = 1; i <= N; i++) read(W[i]);
  root = make_tree(1, N);
  while (Q--) {
    int l, r;
    read(l, r);
    if (l < 0) {
      write(root, 1, N, -l, r);
    } else {
      VALUE all = get(root, 1, N, l, r);
      long long bound = all.sw + 1 >> 1;
      int mid = get_mid(root, 1, N, l, r, bound);
      VALUE vl = get(root, 1, N, l, mid - 1);
      VALUE vr = get(root, 1, N, mid + 1, r);
      int ans = ((+(vl.sw % MOD * X[mid] % MOD - vl.swx + MOD) % MOD +
                  (vr.swx - vr.sw % MOD * X[mid] % MOD + MOD) % MOD -
                  (vl.sw * mid - vl.swi) % MOD - (vr.swi - vr.sw * mid) % MOD) %
                     MOD +
                 MOD) %
                MOD;
      write(ans);
      puts("");
    }
  }
}