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;
void getre() {
  int x = 0;
  printf("%d\n", 1 / x);
}
void gettle() {
  int res = 1;
  while (1) res <<= 1;
  printf("%d\n", res);
}
template <typename T, typename S>
inline bool upmin(T &a, const S &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename S>
inline bool upmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename N, typename PN>
inline N flo(N a, PN b) {
  return a >= 0 ? a / b : -((-a - 1) / b) - 1;
}
template <typename N, typename PN>
inline N cei(N a, PN b) {
  return a > 0 ? (a - 1) / b + 1 : -(-a / b);
}
template <typename N>
N gcd(N a, N b) {
  return b ? gcd(b, a % b) : a;
}
template <typename N>
inline int sgn(N a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
inline void gn(long long &x) {
  int sg = 1;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  c == '-' ? (sg = -1, x = 0) : (x = c - '0');
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= sg;
}
inline void gn(int &x) {
  long long t;
  gn(t);
  x = t;
}
inline void gn(unsigned long long &x) {
  long long t;
  gn(t);
  x = t;
}
inline void gn(double &x) {
  double t;
  scanf("%lf", &t);
  x = t;
}
inline void gn(long double &x) {
  double t;
  scanf("%lf", &t);
  x = t;
}
inline void gs(char *s) { scanf("%s", s); }
inline void gc(char &c) {
  while ((c = getchar()) > 126 || c < 33)
    ;
}
inline void pc(char c) { putchar(c); }
inline long long sqr(long long a) { return a * a; }
inline double sqrf(double a) { return a * a; }
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846264338327950288L;
const double eps = 1e-6;
const int mo = 1e9 + 7;
int n, q;
long long bit[222222];
int bit2[222222];
long long bitque(int x) {
  long long su = 0;
  for (; x; x -= x & -x) su += bit[x];
  return su;
}
int bit2que(int x) {
  int su = 0;
  for (; x; x -= x & -x)
    (((su) = ((su) + (bit2[x])) % mo) < 0 ? (su) += mo : (su));
  return su;
}
void bitupd(int x, long long del) {
  for (; x <= n; x += x & -x) bit[x] += del;
}
void bit2upd(int x, int del) {
  for (; x <= n; x += x & -x)
    (((bit2[x]) = ((bit2[x]) + (del)) % mo) < 0 ? (bit2[x]) += mo : (bit2[x]));
}
int a[222222], w[222222];
int main() {
  gn(n);
  gn(q);
  for (int i = (1), _ed = (n + 1); i < _ed; i++) {
    gn(a[i]);
    a[i] += n - i;
  }
  for (int i = (1), _ed = (n + 1); i < _ed; i++) {
    gn(w[i]);
    bitupd(i, w[i]);
    bit2upd(i, 1ll * a[i] * w[i] % mo);
  }
  while (q--) {
    int x, y;
    gn(x);
    gn(y);
    if (x < 0) {
      x = -x;
      int del = y - w[x];
      w[x] = y;
      bitupd(x, del);
      bit2upd(x, 1ll * a[x] * del % mo);
    } else {
      long long fa = bitque(y);
      long long left = bitque(x - 1);
      long long su = fa - left;
      long long th = (su + 1) / 2;
      long long want = th + left;
      int l = x, r = y;
      while (l <= r) {
        int mid = l + r >> 1;
        long long has = bitque(mid);
        if (has >= want)
          r = mid - 1;
        else
          l = mid + 1;
      }
      int rigsum = ((bit2que(y) - bit2que(l - 1)) -
                    1ll * (bitque(y) - bitque(l - 1)) % mo * a[l]) %
                   mo;
      int lefsum = (-(bit2que(l - 1) - bit2que(x - 1)) +
                    1ll * (bitque(l - 1) - bitque(x - 1)) % mo * a[l]) %
                   mo;
      int ret = ((0ll + rigsum + lefsum) % mo + mo) % mo;
      printf("%d\n", ret);
    }
  }
  return 0;
}