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;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int mod = 1e9 + 7;
const int maxn = 200200;
int n, q, a[maxn];
struct Node {
  int l, r;
  long long val;
} tr[maxn * 4];
long long w[maxn];
struct BIT {
  long long a[maxn * 2];
  void add(int x, long long k) {
    while (x <= 400000) {
      a[x] += k;
      x += (x & -x);
    }
  }
  long long ask(int x) {
    long long ret = 0;
    while (x) {
      ret += a[x];
      x -= (x & -x);
    }
    return ret;
  }
  long long sum(int l, int r) { return ask(r) - ask(l - 1); }
} lzt1, lzt2;
void build(int i, int l, int r) {
  tr[i].l = l, tr[i].r = r;
  if (l == r) {
    tr[i].val = w[tr[i].l];
    return;
  }
  int md = (l + r) >> 1;
  build((i << 1), l, md);
  build((i << 1 | 1), md + 1, r);
  tr[i].val = tr[(i << 1)].val + tr[(i << 1 | 1)].val;
}
void update(int i, int l, int r) {
  if (tr[i].l == tr[i].r) {
    tr[i].val = r;
    return;
  }
  int md = (tr[i].l + tr[i].r) >> 1;
  if (md >= l)
    update((i << 1), l, r);
  else
    update((i << 1 | 1), l, r);
  tr[i].val = tr[(i << 1)].val + tr[(i << 1 | 1)].val;
}
int fnd(int i, long long o) {
  if (tr[i].l == tr[i].r) {
    return tr[i].l;
  }
  if (tr[(i << 1)].val >= o)
    return fnd((i << 1), o);
  else
    return fnd((i << 1 | 1), o - tr[(i << 1)].val);
}
int main() {
  n = read(), q = read();
  for (register int i = (int)(1); i <= (int)(n); i++) a[i] = read() - i + 1;
  for (register int i = (int)(1); i <= (int)(n); i++) {
    w[i] = read();
    lzt1.add(i, w[i]);
    lzt2.add(i, w[i] * 1ll * a[i] % mod);
  }
  build(1, 1, n);
  while (q--) {
    int l = read(), r = read();
    if (l < 0) {
      l = -l;
      lzt1.add(l, -w[l]);
      lzt2.add(l, -(w[l] * 1ll * a[l]) % mod);
      w[l] = r;
      lzt1.add(l, w[l]);
      lzt2.add(l, w[l] * 1ll * a[l] % mod);
      update(1, l, r);
    } else {
      long long L = lzt1.ask(l - 1), R = lzt1.ask(r);
      int k = fnd(1, (L + R + 1) >> 1ll);
      long long ans = a[k] * 1ll * (lzt1.sum(l, k - 1) % mod) % mod -
                      lzt2.sum(l, k - 1) % mod + lzt2.sum(k + 1, r) % mod -
                      a[k] * 1ll * (lzt1.sum(k + 1, r) % mod) % mod;
      ans = (ans % mod + mod) % mod;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}