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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
  Finish_read = 0;
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    if (ch == EOF) return;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
  Finish_read = 1;
}
template <class T>
inline void print(T x) {
  if (x / 10 != 0) print(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void writeln(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
  putchar('\n');
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-');
  x = abs(x);
  print(x);
}
const int maxn = 200005, mod = 1e9 + 7;
int n, m, a[maxn], w[maxn], x, y;
namespace {
inline int Add(const int &x, const int &y) {
  int res = x + y;
  return res >= mod ? res - mod : res;
}
inline int Sub(const int &x, const int &y) {
  int res = x - y;
  return res < 0 ? res + mod : res;
}
inline int Mul(const int &x, const int &y) { return 1ll * x * y % mod; }
}  // namespace
namespace Segtree {
int sq[maxn << 2];
long long sum[maxn << 2];
inline void Build(int o, int l, int r) {
  if (l == r) {
    sq[o] = Mul(a[l], w[l]), sum[o] = w[l];
    return;
  }
  Build(o << 1, l, ((l + r) >> 1)), Build(o << 1 | 1, ((l + r) >> 1) + 1, r);
  sq[o] = Add(sq[o << 1], sq[o << 1 | 1]),
  sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
inline void Modify(int o, int l, int r, int x, int y) {
  if (l == r) {
    sq[o] = Mul(a[l], y), sum[o] = y;
    return;
  }
  if (x <= ((l + r) >> 1))
    Modify(o << 1, l, ((l + r) >> 1), x, y);
  else
    Modify(o << 1 | 1, ((l + r) >> 1) + 1, r, x, y);
  sq[o] = Add(sq[o << 1], sq[o << 1 | 1]),
  sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
inline int Querys(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sq[o];
  int res = 0;
  if (ql <= ((l + r) >> 1))
    res = Add(res, Querys(o << 1, l, ((l + r) >> 1), ql, qr));
  if (qr > ((l + r) >> 1))
    res = Add(res, Querys(o << 1 | 1, ((l + r) >> 1) + 1, r, ql, qr));
  return res;
}
inline long long Querya(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[o];
  long long res = 0;
  if (ql <= ((l + r) >> 1)) res += Querya(o << 1, l, ((l + r) >> 1), ql, qr);
  if (qr > ((l + r) >> 1))
    res += Querya(o << 1 | 1, ((l + r) >> 1) + 1, r, ql, qr);
  return res;
}
inline void Printa(int o, int l, int r) {
  cout << "Interval:" << l << " -> " << r << " Sum:" << sum[o] << endl;
  if (l == r) return;
  Printa(o << 1, l, ((l + r) >> 1)), Printa(o << 1 | 1, ((l + r) >> 1) + 1, r);
}
}  // namespace Segtree
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) read(a[i]), a[i] -= i;
  for (int i = 1; i <= n; ++i) read(w[i]);
  Segtree::Build(1, 1, n);
  while (m--) {
    read(x), read(y);
    if (x < 0)
      Segtree::Modify(1, 1, n, -x, y);
    else {
      long long tot = Segtree::Querya(1, 1, n, x, y);
      tot = (tot + 1) >> 1;
      int l = x, r = n, md;
      while (l < r) {
        md = (l + r) >> 1;
        if (Segtree::Querya(1, 1, n, x, md) < tot)
          l = md + 1;
        else
          r = md;
      }
      int p = l;
      printf("%d\n",
             Add(Sub(Mul(Segtree::Querya(1, 1, n, x, p) % mod, a[p]),
                     Segtree::Querys(1, 1, n, x, p)),
                 Sub(Segtree::Querys(1, 1, n, p + 1, y),
                     Mul(Segtree::Querya(1, 1, n, p + 1, y) % mod, a[p]))));
    }
  }
}