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;
int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar();
  return v * f;
}
const int N = 201000;
const long long MOD = 1e9 + 7;
int n, q;
long long t1[N << 2], t2[N << 2], a[N], t, w[N];
void modify(int u, int l, int r, int p, long long v) {
  if (l == r) {
    t1[u] = v;
    t2[u] = v * (a[p] - p) % MOD;
    return;
  }
  int mid = l + r >> 1;
  if (p <= mid)
    modify(u << 1, l, mid, p, v);
  else
    modify(u << 1 | 1, mid + 1, r, p, v);
  t1[u] = t1[u << 1] + t1[u << 1 | 1];
  t2[u] = (t2[u << 1] + t2[u << 1 | 1]) % MOD;
}
long long query(int u, int l, int r, int L, int R) {
  if (L > R) return 0;
  if (L <= l && r <= R) return t1[u];
  int mid = l + r >> 1;
  long long re = 0;
  if (L <= mid) re += query(u << 1, l, mid, L, R);
  if (mid < R) re += query(u << 1 | 1, mid + 1, r, L, R);
  return re;
}
long long query2(int u, int l, int r, int L, int R) {
  if (L > R) return 0;
  if (L <= l && r <= R) return t2[u];
  int mid = l + r >> 1;
  long long re = 0;
  if (L <= mid) re += query2(u << 1, l, mid, L, R);
  if (mid < R) re += query2(u << 1 | 1, mid + 1, r, L, R);
  return re % MOD;
}
int find2(int u, int l, int r) {
  if (l == r) return l;
  int mid = l + r >> 1;
  if (t - t1[u << 1] > 0) {
    t -= t1[u << 1];
    return find2(u << 1 | 1, mid + 1, r);
  } else {
    return find2(u << 1, l, mid);
  }
}
int find(int u, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    if (t - t1[u] > 0)
      t -= t1[u];
    else
      return find2(u, l, r);
    return 0;
  }
  int mid = l + r >> 1;
  if (L <= mid) {
    int tmp = find(u << 1, l, mid, L, R);
    if (tmp) return tmp;
  }
  if (mid < R) {
    int tmp = find(u << 1 | 1, mid + 1, r, L, R);
    if (tmp) return tmp;
  }
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) w[i] = read();
  for (int i = 1; i <= n; i++) modify(1, 1, n, i, w[i]);
  while (q--) {
    int u = read(), v = read();
    if (u < 0) {
      u = -u;
      modify(1, 1, n, u, v);
    } else {
      t = query(1, 1, n, u, v);
      t = (t + 1) / 2;
      int p = find(1, 1, n, u, v);
      long long x = (a[p] - p) % MOD;
      printf("%lld\n",
             (query(1, 1, n, u, p) % MOD * x % MOD - query2(1, 1, n, u, p) -
              query(1, 1, n, p + 1, v) % MOD * x % MOD +
              query2(1, 1, n, p + 1, v) + MOD + MOD) %
                 MOD);
    }
  }
}