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 void read(T &x) {
  x = 0;
  int fu = 1;
  char c = getchar();
  while (c > 57 || c < 48) {
    if (c == 45) fu = -1;
    c = getchar();
  }
  while (c <= 57 && c >= 48) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  x *= fu;
}
template <typename T>
inline void fprint(T x) {
  if (x < 0) putchar(45), x = -x;
  if (x > 9) fprint(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void fprint(T x, char ch) {
  fprint(x);
  putchar(ch);
}
inline char next_char() {
  char ch = getchar();
  while (ch == 9 || ch == 10 || ch == 32) ch = getchar();
  return ch;
}
const long long MOD = 1000000007;
int n, m;
long long a[200005], w[200005];
long long c1[200005], c2[200005];
inline int lowbit(int x) { return x & -x; }
inline long long query(long long c[], int x, bool type) {
  long long ret = 0;
  for (; x; x -= lowbit(x)) ret += c[x], (type) && (ret = (ret + MOD) % MOD);
  return ret;
}
inline void modify(long long c[], int x, long long y, bool type) {
  for (; x <= n; x += lowbit(x))
    c[x] += y, (type) && (c[x] = (c[x] + MOD) % MOD);
}
int main() {
  read(n);
  read(m);
  for (register int i = 1; i <= n; i++) read(a[i]);
  for (register int i = 1; i <= n; i++)
    read(w[i]), modify(c1, i, w[i], 0),
        modify(c2, i, (a[i] - i) * w[i] % MOD, 1);
  while (m--) {
    int l, r;
    read(l);
    read(r);
    if (l < 0) {
      l = -l;
      modify(c1, l, -w[l], 0);
      modify(c2, l, ((r - w[l]) * (a[l] - l) % MOD + MOD) % MOD, 1);
      w[l] = r;
      modify(c1, l, r, 0);
    } else {
      int L = l, R = r, pos = l;
      long long ll = query(c1, l - 1, 0), rr = query(c1, r, 0);
      long long tot = rr - ll;
      tot = (tot >> 1) + 1;
      while (L <= R) {
        int mid = (L + R) >> 1;
        if (query(c1, mid, 0) - ll >= tot)
          pos = mid, R = mid - 1;
        else
          L = mid + 1;
      }
      long long res1 = query(c1, pos, 0), res2 = query(c1, pos - 1, 0);
      rr %= MOD, ll %= MOD, res1 %= MOD, res2 %= MOD;
      long long num = (a[pos] - (pos - l + 1) + MOD) % MOD;
      long long ans = num * (res2 - ll) % MOD;
      ans =
          ((ans + 1ll * (l - 1) * (rr - res1 - (res2 - ll))) % MOD + MOD) % MOD;
      ans = ((ans - query(c2, pos - 1, 1) + query(c2, l - 1, 1)) % MOD + MOD) %
            MOD;
      ans = (ans + query(c2, r, 1) - query(c2, pos, 1) + MOD) % MOD;
      ans = (ans - ((rr - res1) * num % MOD) + MOD) % MOD;
      fprint(ans, 10);
    }
  }
}