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 = 2e6 + 100;
long long mod = 1e9 + 7;
int n, q;
long long a[maxn], w[maxn], x, y;
struct semergetree {
  long long sum[maxn * 4], v[maxn];
  inline void push_up(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
  inline void build(int rt, int l, int r) {
    if (l == r) {
      sum[rt] = v[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    push_up(rt);
  }
  void update(int rt, int l, int r, int pos, long long val) {
    if (l == r) {
      sum[rt] = val;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      update(rt << 1, l, mid, pos, val);
    else
      update(rt << 1 | 1, mid + 1, r, pos, val);
    push_up(rt);
  }
  long long query(int rt, int l, int r, int L, int R) {
    if (L <= l && r <= R) {
      return sum[rt];
    }
    int mid = (l + r) >> 1;
    if (R <= mid)
      return query(rt << 1, l, mid, L, R);
    else if (L > mid)
      return query(rt << 1 | 1, mid + 1, r, L, R);
    else
      return query(rt << 1, l, mid, L, mid) +
             query(rt << 1 | 1, mid + 1, r, mid + 1, R);
  }
  int find_k(int rt, int l, int r, int L, int R, long long val) {
    if (l == r) {
      return l;
    }
    int mid = (l + r) >> 1;
    if (R <= mid) return find_k(rt << 1, l, mid, L, R, val);
    if (L > mid) return find_k(rt << 1 | 1, mid + 1, r, L, R, val);
    long long res = query(rt << 1, l, mid, L, mid);
    if (res >= val)
      return find_k(rt << 1, l, mid, L, mid, val);
    else
      return find_k(rt << 1 | 1, mid + 1, r, mid + 1, R, val - res);
  }
} T1, T2;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), a[i] -= i;
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]);
  for (int i = 1; i <= n; i++) T1.v[i] = w[i], T2.v[i] = (a[i] * w[i]) % mod;
  T1.build(1, 1, n);
  T2.build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    scanf("%lld%lld", &x, &y);
    if (x < 0) {
      x = -x;
      w[x] = y;
      T1.update(1, 1, n, x, w[x]);
      T2.update(1, 1, n, x, (a[x] * w[x]) % mod);
    } else {
      long long sum = T1.query(1, 1, n, x, y);
      int pos = T1.find_k(1, 1, n, x, y, (sum + 1) >> 1);
      long long ans = 0;
      ans = ans +
            (a[pos] * (T1.query(1, 1, n, x, pos) - T1.query(1, 1, n, pos, y)) %
             mod) %
                mod;
      while (ans < 0) ans += mod;
      ans = ans - (T2.query(1, 1, n, x, pos) - T2.query(1, 1, n, pos, y)) % mod;
      ans %= mod;
      while (ans < 0) ans += mod;
      cout << ans << endl;
    }
  }
  return 0;
}