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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void pre() {}
void solve() {}
const long long mod = 1e9 + 7;
struct FT {
  vector<long long> s;
  FT(int n) : s(n, 0) {}
  void update(int pos, long long dif) {
    for (; pos < (int)(s).size(); pos |= pos + 1) s[pos] += dif;
  }
  long long query(int pos) {
    long long res = 0;
    for (; pos > 0; pos &= pos - 1) res += s[pos - 1];
    return res;
  }
};
struct Node {
  Node *l = 0, *r = 0;
  int lo, hi;
  long long madd = 0, val = 0, d;
  Node(vector<int>& v, int lo, int hi) : lo(lo), hi(hi) {
    if (lo + 1 < hi) {
      int mid = lo + (hi - lo) / 2;
      l = new Node(v, lo, mid);
      r = new Node(v, mid, hi);
      d = (l->d + r->d) % mod;
    } else if (lo + 1 == hi)
      d = v[lo];
  }
  long long query(int L, int R) {
    if (R <= lo || hi <= L) return 0;
    if (L <= lo && hi <= R) return val;
    push();
    return (l->query(L, R) + r->query(L, R)) % mod;
  }
  void add(int L, int R, long long x) {
    if (R <= lo || hi <= L) return;
    if (L <= lo && hi <= R) {
      madd += x;
      val += x % mod * d % mod;
      madd %= mod;
      val %= mod;
    } else {
      push(), l->add(L, R, x), r->add(L, R, x);
      val = (l->val + r->val) % mod;
    }
  }
  void push() {
    if (madd)
      l->add(lo, hi, madd), r->add(lo, hi, madd), madd = 0,
                                                  val = (l->val + r->val) % mod;
  }
};
long long a[200009], w[200009];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pre();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < (n); ++i) cin >> a[i];
  for (int i = 0; i < (n); ++i) cin >> w[i];
  FT T1(n + 3), T2(n + 3), T3(n + 3), T4(n + 3);
  long long sum = 0;
  vector<int> d(n - 1);
  for (int i = 0; i < (n - 1); ++i)
    d[i] = a[i + 1] - a[i] - 1, T4.update(i, d[i]);
  Node N1(d, 0, n - 1), N2(d, 0, n - 1);
  for (int i = 0; i < (n - 1); ++i) {
    sum += w[i];
    sum %= mod;
    T1.update(i, 1ll * (a[i + 1] - a[i] - 1) * sum % mod);
  }
  sum = 0;
  for (int i = n - 2; i >= (0); --i) {
    sum += w[i + 1];
    sum %= mod;
    T2.update(i, 1ll * (a[i + 1] - a[i] - 1) * sum % mod);
  }
  for (int i = 0; i < (n); ++i) {
    T3.update(i, w[i]);
  }
  for (int i = 0; i < (q); ++i) {
    int l, r;
    cin >> l >> r;
    if (l < 0) {
      int ix = -l;
      ix--;
      long long dif = r - w[ix];
      dif %= mod;
      T3.update(ix, r - w[ix]);
      w[ix] = r;
      if (ix > 0) {
        N2.add(0, ix, dif);
      }
      if (ix < n - 1) {
        N1.add(ix, n, dif);
      }
    } else {
      l--, r--;
      int lo = l, hi = r;
      while (lo < hi) {
        int m = (lo + hi) / 2;
        if (T3.query(r + 1) - T3.query(m + 1) >=
            T3.query(m + 1) - T3.query(l)) {
          lo = m + 1;
        } else
          hi = m;
      }
      long long ans =
          T2.query(r) - T2.query(lo) + T1.query(lo) - T1.query(l) -
          (T4.query(lo) - T4.query(l)) % mod * (T3.query(l) % mod) % mod -
          (T4.query(r) - T4.query(lo)) % mod *
              ((T3.query(n) - T3.query(r + 1)) % mod) % mod;
      ans += N1.query(l, lo) + N2.query(lo, r);
      ans = (ans % mod + mod) % mod;
      cout << ans << '\n';
    }
  }
  return 0;
}