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 N = 202020, P = 1e9 + 7;
int n, m;
int a[N], w[N];
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
struct Seg {
  static const int N = ::N << 2;
  long long sum[N];
  int num[N];
  vector<pair<pair<int, int>, int> > v;
  void up(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    num[rt] = add(num[rt << 1], num[rt << 1 | 1]);
  }
  void build(int l, int r, int rt) {
    if (l == r) {
      sum[rt] = w[l];
      num[rt] = mul(w[l], sub(l, a[l]));
      return;
    }
    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    up(rt);
  }
  void upd(int p, int c, int l, int r, int rt) {
    if (l == r) {
      sum[rt] = w[l] = c;
      num[rt] = mul(c, sub(l, a[l]));
      return;
    }
    int mid = l + r >> 1;
    if (p <= mid)
      upd(p, c, l, mid, rt << 1);
    else
      upd(p, c, mid + 1, r, rt << 1 | 1);
    up(rt);
  }
  pair<int, int> qry2(int L, int R, int l, int r, int rt) {
    if (L > R) return make_pair(0, 0);
    if (L <= l && r <= R) return make_pair(num[rt], sum[rt] % P);
    int mid = l + r >> 1, res = 0, ans = 0;
    if (L <= mid) {
      auto u = qry2(L, R, l, mid, rt << 1);
      res = add(res, u.first);
      ans = add(ans, u.second);
    }
    if (R > mid) {
      auto u = qry2(L, R, mid + 1, r, rt << 1 | 1);
      res = add(res, u.first);
      ans = add(ans, u.second);
    }
    return make_pair(res, ans);
  }
  long long qry(int L, int R, int l, int r, int rt) {
    if (rt == 1) v.clear();
    if (L <= l && r <= R) {
      v.push_back(make_pair(make_pair(l, r), rt));
      return sum[rt];
    }
    int mid = l + r >> 1;
    long long res = 0;
    if (L <= mid) res += qry(L, R, l, mid, rt << 1);
    if (R > mid) res += qry(L, R, mid + 1, r, rt << 1 | 1);
    return res;
  }
  int qry(int L, int R) {
    long long c = qry(L, R, 1, n, 1), s = 0;
    for (auto u : v) {
      s += sum[u.second];
      if (s >= c - s) {
        s -= sum[u.second];
        int l = u.first.first, r = u.first.second, rt = u.second;
        while (l < r) {
          int mid = l + r >> 1;
          if (s + sum[rt << 1] >= c - s - sum[rt << 1]) {
            r = mid;
            rt = rt << 1;
          } else {
            s += sum[rt << 1];
            l = mid + 1;
            rt = rt << 1 | 1;
          }
        }
        return l;
      }
    }
  }
} seg;
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  cin >> n >> m;
  for (int i = (1); i < (n + 1); ++i) cin >> a[i];
  for (int i = (1); i < (n + 1); ++i) cin >> w[i];
  seg.build(1, n, 1);
  while (m--) {
    int x, y;
    cin >> x >> y;
    if (x > 0) {
      int k = seg.qry(x, y);
      auto L = seg.qry2(x, k - 1, 1, n, 1);
      auto R = seg.qry2(k + 1, y, 1, n, 1);
      int ans = add(mul(sub(a[k], k), L.second), L.first);
      int res = sub(mul(sub(k, a[k]), R.second), R.first);
      cout << add(ans, res) << "\n";
    } else {
      seg.upd(-x, y, 1, n, 1);
    }
  }
  return 0;
}