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 int64 = long long;
constexpr int Maxn = 200000;
constexpr int p = 1000000007;
template <class _Tp1, class _Tp2>
inline void inc(_Tp1& x, const _Tp2& y) {
  x += y;
  (p <= x) && (x -= p);
}
template <class _Tp1, class _Tp2>
inline void dec(_Tp1& x, const _Tp2& y) {
  x -= y;
  (x < 0) && (x += p);
}
template <class _Tp>
inline int md(const _Tp& v) {
  return v < 0 ? ((0 <= p + v) ? (p + v) : (v % p + p)) : (v < p) ? v : (v % p);
}
namespace IOManager {
constexpr int FILESZ(131072);
unsigned char buf[FILESZ];
const unsigned char *ibuf = buf, *tbuf = buf;
struct IOManager {
  inline unsigned char gc() {
    return (ibuf == tbuf) &&
                   (tbuf = (ibuf = buf) + fread(buf, 1, FILESZ, stdin),
                    ibuf == tbuf)
               ? EOF
               : *ibuf++;
  }
  template <class _Tp>
  inline operator _Tp() {
    _Tp s = 0u;
    unsigned char c = gc(), w = 0;
    for (; c < 48; c = gc()) (c == 45) && (w = 1);
    for (; c > 47; c = gc()) s = (_Tp)(s * 10u + c - 48u);
    return w ? -s : s;
  }
};
}  // namespace IOManager
IOManager::IOManager io;
const int n = io;
int64 bit[Maxn + 1];
inline void update(const int& x, const int& d) {
  for (int i = x; i <= n; i += i & -i) bit[i] += d;
}
inline int64 sum(const int& x) {
  int64 s = 0;
  for (int i = x; i; i -= i & -i) s += bit[i];
  return s;
}
int bitp[Maxn + 1];
inline void updatep(const int& x, const int& d) {
  for (int i = x; i <= n; i += i & -i) inc(bitp[i], d);
}
inline int sump(const int& x) {
  int64 s = 0;
  for (int i = x; i; i -= i & -i) s += bitp[i];
  return s % p;
}
int a[Maxn + 1], w[Maxn + 1];
int main() {
  int q = io;
  for (int i = 1; i <= n; ++i) a[i] = (int)io - i;
  for (int i = 1; i <= n; ++i)
    w[i] = io, update(i, w[i]), updatep(i, (int64)w[i] * a[i] % p);
  int64 sl, sm, sr;
  for (int x, y, l, m, r; q; --q)
    if (x = io, y = io, x < 0) {
      int d = y - w[x = -x];
      w[x] = y;
      update(x, d);
      updatep(x, md((int64)d * a[x]));
    } else {
      l = x;
      r = y;
      sl = sum(l - 1);
      sr = sum(r) - sl;
      for (; l < r;)
        ((sum(m = (l + r) >> 1) - sl) * 2 < sr) ? (l = m + 1) : (r = m);
      sm = sum(m = l);
      sr += sl;
      printf("%d\n", md((int64)(sump(y) - sump(m) * 2ll +
                                (int64)(sm - sr + sm - sl) % p * a[m] % p +
                                sump(x - 1))));
    }
  return 0;
}