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 INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
long long c[N], sum[N << 2], w[N], sum_fn[N << 2];
int n, a[N];
int lowbit(int x) { return x & -x; }
void upd(int x, int val) {
  for (; x <= n; x += lowbit(x)) c[x] += val;
}
long long get_sum(int x) {
  long long ans = 0;
  for (; x; x -= lowbit(x)) ans += c[x];
  return ans;
}
void build(int l, int r, int rt) {
  if (l == r) {
    sum[rt] = w[l] * l % mod;
    sum_fn[rt] = w[l] * a[l] % mod;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, (rt << 1));
  build(mid + 1, r, (rt << 1 | 1));
  sum[rt] = (sum[(rt << 1)] + sum[(rt << 1 | 1)]) % mod;
  sum_fn[rt] = (sum_fn[(rt << 1)] + sum_fn[(rt << 1 | 1)]) % mod;
}
void update(int l, int r, int x, int val, int rt) {
  if (l == r) {
    sum[rt] = val * 1ll * l % mod;
    sum_fn[rt] = val * 1ll * a[l] % mod;
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    update(l, mid, x, val, (rt << 1));
  else
    update(mid + 1, r, x, val, (rt << 1 | 1));
  sum[rt] = (sum[(rt << 1)] + sum[(rt << 1 | 1)]) % mod;
  sum_fn[rt] = (sum_fn[(rt << 1)] + sum_fn[(rt << 1 | 1)]) % mod;
}
pair<long long, long long> query(int l, int r, int L, int R, int rt) {
  if (L <= l && r <= R) return make_pair(sum[rt], sum_fn[rt]);
  int mid = l + r >> 1;
  pair<long long, long long> ans = make_pair(0ll, 0ll);
  if (L <= mid) {
    pair<long long, long long> res = query(l, mid, L, R, (rt << 1));
    ans.first = (ans.first + res.first) % mod;
    ans.second = (ans.second + res.second) % mod;
  }
  if (mid < R) {
    pair<long long, long long> res = query(mid + 1, r, L, R, (rt << 1 | 1));
    ans.first = (ans.first + res.first) % mod;
    ans.second = (ans.second + res.second) % mod;
  }
  return ans;
}
int main() {
  int q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]), upd(i, w[i]);
  build(1, n, 1);
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      x = -x;
      update(1, n, x, y, 1);
      upd(x, y - w[x]);
      w[x] = y;
    } else {
      long long pr = get_sum(x - 1), sb = get_sum(y);
      int l = x, r = y, ans = -1;
      while (l <= r) {
        int mid = l + r >> 1;
        long long res = get_sum(mid);
        if (res - pr >= sb - res) {
          ans = mid;
          r = mid - 1;
        } else
          l = mid + 1;
      }
      long long su = 0;
      long long w_ans = get_sum(ans);
      if (ans < y) {
        pair<long long, long long> p = query(1, n, ans + 1, y, 1);
        su += (p.second - p.first -
               (sb - w_ans) % mod * 1ll * (a[ans] - ans) % mod) %
              mod;
      }
      if (ans >= x) {
        pair<long long, long long> p = query(1, n, x, ans, 1);
        su += (p.first - p.second +
               (w_ans - pr) % mod * 1ll * (a[ans] - ans) % mod) %
              mod;
      }
      su %= mod;
      su = (su + mod) % mod;
      printf("%lld\n", su);
    }
  }
  return 0;
}