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;
template <typename T>
inline void read(T& x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T, typename... U>
inline void read(T& x, U&... y) {
  read(x), read(y...);
}
const int N = 2e5 + 10, P = 1e9 + 7;
int n, Q;
int A[N], W[N];
long long X[N], S[N], G[N];
void Modify(int pos, int val) {
  while (pos <= n) {
    X[pos] += val;
    pos += (pos & -pos);
  }
}
long long Sum(int pos) {
  long long res = 0;
  while (pos) {
    res += X[pos];
    pos -= (pos & -pos);
  }
  return res;
}
void ADD(int pos, int val) {
  while (pos <= n) {
    S[pos] += val;
    if (S[pos] >= P) S[pos] -= P;
    pos += (pos & -pos);
  }
}
long long Get(int pos) {
  long long res = 0;
  while (pos) {
    res += S[pos];
    if (res >= P) res -= P;
    pos -= (pos & -pos);
  }
  return res;
}
void Update(int pos, int val) {
  while (pos <= n) {
    G[pos] += val;
    if (S[pos] >= P) G[pos] -= P;
    pos += (pos & -pos);
  }
}
long long Calc(int pos) {
  long long res = 0;
  while (pos) {
    res += G[pos];
    if (res >= P) res -= P;
    pos -= (pos & -pos);
  }
  return res;
}
int Find(int l, int r, long long x, long long Sl) {
  if (l == r) return l;
  int mid = l + r >> 1;
  long long t = Sum(mid) - Sl;
  if (t >= x - t) return Find(l, mid, x, Sl);
  return Find(mid + 1, r, x, Sl);
}
int Query(int l, int r) {
  if (l == r) return 0;
  int p = Find(l, r, Sum(r) - Sum(l - 1), Sum(l - 1));
  long long SUM = (Sum(p) - Sum(l - 1)) % P, S1 = 1ll * A[p] * SUM % P,
            S2 = 1ll * p * SUM % P, S3 = (Calc(p) - Calc(l - 1)) % P,
            S4 = (Get(p) - Get(l - 1)) % P;
  long long t1 = ((S1 - S2 + S3 - S4) % P + P) % P;
  SUM = (Sum(r) - Sum(p)) % P, S1 = (Get(r) - Get(p)) % P,
  S2 = 1ll * A[p] * SUM % P, S3 = (Calc(r) - Calc(p)) % P,
  S4 = 1ll * p * SUM % P;
  long long t2 = ((S1 - S2 - S3 + S4) % P + P) % P;
  return (t1 + t2) % P;
}
int main() {
  read(n, Q);
  for (int i = 1; i <= n; ++i) read(A[i]);
  for (int i = 1; i <= n; ++i)
    read(W[i]), Modify(i, W[i]), ADD(i, 1ll * W[i] * A[i] % P),
        Update(i, 1ll * i * W[i] % P);
  while (Q--) {
    int l, r;
    read(l, r);
    if (l < 0)
      Modify(-l, r - W[-l]), ADD(-l, (1ll * (r - W[-l]) * A[-l] % P + P) % P),
          Update(-l, (1ll * (r - W[-l]) * (-l) % P + P) % P), W[-l] = r;
    else
      printf("%d\n", Query(l, r));
  }
  return 0;
}