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;
#pragma GCC optimize("Ofast")
const int MAXN = 2e5 + 20;
int a[MAXN], w[MAXN], n;
const int P = 1e9 + 7;
inline void MOD(int& p) {
  while (p < 0) p += P;
  while (p >= P) p -= P;
}
struct BIT1 {
  long long bit[MAXN];
  inline void modify(int w, long long v) {
    while (w <= n) {
      bit[w] += v;
      w += (w & (-w));
    }
  }
  inline long long sum(int w) {
    long long res = 0;
    while (w) {
      res += bit[w];
      w -= (w & (-w));
    }
    return res;
  }
  inline long long query(int l, int r) { return sum(r) - sum(l - 1); }
} BitW;
struct BIT2 {
  int bit[MAXN];
  inline void modify(int w, int v) {
    while (w <= n) {
      MOD(bit[w] += v);
      w += (w & (-w));
    }
  }
  inline int sum(int w) {
    MOD(w);
    long long res = 0;
    while (w) {
      res += bit[w];
      w -= (w & (-w));
    }
    return res % P;
  }
  inline int query(int l, int r) { return (sum(r) - sum(l - 1) + P) % P; }
} BitS;
inline int getAns(int l, int r) {
  int L = l, R = r;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (BitW.query(l, mid) >= BitW.query(mid + 1, r)) {
      R = mid - 1;
    } else {
      L = mid + 1;
    }
  }
  ++R;
  long long res = 0;
  res = -BitS.query(l, R) + BitW.query(l, R) % P * abs(a[R] - R) % P;
  res = (res % P + P) % P;
  res -= -BitS.query(R, r) + BitW.query(R, r) % P * abs(a[R] - R) % P;
  res = (res % P + P) % P;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int q;
  cin >> n >> q;
  for (register int i = (1); i <= (n); ++i) {
    cin >> a[i];
  }
  for (register int i = (1); i <= (n); ++i) {
    cin >> w[i];
    BitW.modify(i, w[i]);
    BitS.modify(i, 1ll * w[i] * (a[i] - i) % P);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (l < 0) {
      l = -l;
      BitW.modify(l, -w[l]);
      BitS.modify(l, -1ll * w[l] * (a[l] - l) % P);
      w[l] = r;
      BitW.modify(l, w[l]);
      BitS.modify(l, 1ll * w[l] * (a[l] - l) % P);
    } else {
      cout << getAns(l, r) << '\n';
    }
  }
  return 0;
}