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;
typedef long long LL;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - 48;
    ch = getchar();
  }
  return x * f;
}
template <typename T>
inline bool Max(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool Min(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
const int N = 2e5 + 7, mod = 1e9 + 7;
int n, q;
LL w[N], p[N];
void add(LL &a, LL b) {
  a += b;
  if (a >= mod) a -= mod;
}
LL s[N << 2];
void modify(int p, int l, int r, int o, LL v) {
  s[o] += v;
  if (l == r) return;
  int mid = l + r >> 1;
  if (p <= mid)
    modify(p, l, mid, o << 1, v);
  else
    modify(p, mid + 1, r, o << 1 | 1, v);
}
int query(int l, int r, int o, LL sum) {
  if (l == r) return l;
  int mid = l + r >> 1;
  if (s[o << 1] >= sum) return query(l, mid, o << 1, sum);
  return query(mid + 1, r, o << 1 | 1, sum - s[o << 1]);
}
LL querys(int L, int R, int l, int r, int o) {
  if (L > R) return 0;
  if (L <= l && r <= R) return s[o];
  int mid = l + r >> 1;
  if (R <= mid) return querys(L, R, l, mid, o << 1);
  if (L > mid) return querys(L, R, mid + 1, r, o << 1 | 1);
  return querys(L, R, l, mid, o << 1) + querys(L, R, mid + 1, r, o << 1 | 1);
}
struct BIT {
  LL tr[N];
  int lowbit(int x) { return x & -x; }
  void modify(int x, LL y) {
    for (int i = x; i <= n; i += lowbit(i)) add(tr[i], y);
  }
  LL query(int x) {
    LL res = 0ll;
    for (int i = x; i; i -= lowbit(i)) add(res, tr[i]);
    return res;
  }
  LL q(int l, int r) { return query(r) - query(l - 1); }
} t[2];
int main() {
  n = gi(), q = gi();
  for (int i = 1; i <= n; ++i) p[i] = gi();
  for (int i = 1; i <= n; ++i) w[i] = gi() % mod;
  for (int i = 1; i <= n; ++i) {
    t[0].modify(i, p[i] * w[i] % mod);
    t[1].modify(i, i * w[i] % mod);
    modify(i, 1, n, 1, w[i]);
  }
  while (q--) {
    int x = gi(), y = gi();
    if (x < 0) {
      x = -x;
      LL tt = y - w[x];
      modify(x, 1, n, 1, tt);
      t[0].modify(x, p[x] * tt % mod);
      t[1].modify(x, x * tt % mod);
      w[x] = y;
    } else {
      LL ans = 0;
      int l = x, r = y;
      int pos =
          query(1, n, 1, querys(1, y, 1, n, 1) - (querys(x, y, 1, n, 1) / 2));
      add(ans, ((-t[0].q(l, pos) +
                 querys(l, pos, 1, n, 1) % mod * (p[pos] - pos) % mod +
                 t[1].q(l, pos)) %
                    mod +
                mod) %
                   mod);
      add(ans, ((t[0].q(pos + 1, r) -
                 querys(pos + 1, r, 1, n, 1) % mod * (p[pos] - pos) % mod -
                 t[1].q(pos + 1, r)) %
                    mod +
                mod) %
                   mod);
      printf("%I64d\n", ans);
    }
  }
  return 0;
}