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;
const int maxn = 2e5 + 5;
const int P = 1e9 + 7;
int n, m, a[maxn], w[maxn];
inline int read() {
  char ch = getchar();
  int u = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    u = u * 10 + ch - 48;
    ch = getchar();
  }
  return u * f;
}
inline int pls(int x, int y) {
  x += y;
  return x >= P ? x - P : x;
}
inline int dec(int x, int y) {
  x -= y;
  return x < 0 ? x + P : x;
}
inline int mul(int x, int y) {
  long long rnt = 1ll * x * y;
  return (int)(rnt >= P ? rnt % P : rnt);
}
struct BIT {
  long long C[maxn];
  inline void modify(int p, int x) {
    while (p <= n) {
      C[p] += x;
      p += (-p & p);
    }
  }
  inline long long query(int p) {
    long long rnt = 0;
    while (p) {
      rnt += C[p];
      p -= (-p & p);
    }
    return rnt;
  }
} T;
struct _BIT {
  int C[maxn];
  inline void modify(int p, int x) {
    while (p <= n) {
      C[p] = pls(C[p], x);
      p += (-p & p);
    }
  }
  inline int query(int p) {
    int rnt = 0;
    while (p) {
      rnt = pls(rnt, C[p]);
      p -= (-p & p);
    }
    return rnt;
  }
} T0;
inline int getpos(int l, int r) {
  long long S = T.query(r) - T.query(l - 1), tmp = 0;
  int mid = (l + r) / 2, ans = 0, L = l;
  while (l <= r) {
    mid = (l + r) / 2;
    tmp = T.query(mid) - T.query(L - 1);
    if (tmp * 2 >= S)
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  return ans;
}
int main() {
  n = read();
  m = read();
  for (register int i = 1; i <= n; i++) a[i] = read() - i;
  for (register int i = 1; i <= n; i++) {
    w[i] = read();
    T.modify(i, w[i]);
    T0.modify(i, mul(a[i], w[i]));
  }
  for (register int i = 1; i <= m; i++) {
    int l = read(), r = read();
    if (l < 0) {
      T.modify(-l, r - w[-l]);
      T0.modify(-l, dec(mul(a[-l], r), mul(a[-l], w[-l])));
      w[-l] = r;
    } else {
      int pos = getpos(l, r);
      int S0 = (T.query(pos) - T.query(l - 1) + P) % P;
      int S1 = (T.query(r) - T.query(pos) + P) % P;
      int P1 = dec(mul(S0, a[pos]), dec(T0.query(pos), T0.query(l - 1)));
      int P2 = dec(dec(T0.query(r), T0.query(pos)), mul(S1, a[pos]));
      printf("%d\n", pls(P1, P2));
    }
  }
  return 0;
}