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;
using ll = long long int;
template <typename T>
ostream& operator+(ostream& out, const vector<T>& vec) {
  for (const auto& x : vec) {
    out << x << " ";
  }
  out << "\n";
  return out;
}
template <typename T>
ostream& operator*(ostream& out, const vector<T>& vec) {
  for (const auto& x : vec) {
    out + x;
  }
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& vec) {
  for (auto& x : vec) {
    in >> x;
  }
  return in;
}
template <typename T>
struct segment_tree {
  int n;
  vector<T> segtree;
  function<T(const T&, const T&)> join;
  T base = T();
  segment_tree(int n, function<T(const T&, const T&)> join, T base = T())
      : n(n), join(join), base(base) {
    segtree.resize(n << 1, base);
  }
  void build() {
    for (int i = n - 1; i > 0; --i) {
      segtree[i] = join(segtree[(i << 1)], segtree[(i << 1) | 1]);
    }
  }
  segment_tree(vector<T> seq, function<T(const T&, const T&)> join,
               T base = T())
      : n(seq.size()), join(join), base(base) {
    segtree.resize(n << 1, base);
    for (int i = 0; i < n; i++) {
      segtree[n + i] = seq[i];
    }
    build();
  }
  void calc(int pos) {
    pos >>= 1;
    while (pos) {
      segtree[pos] = join(segtree[pos << 1], segtree[(pos << 1) | 1]);
      pos >>= 1;
    }
  }
  void set(int pos, T val) {
    pos += n;
    segtree[pos] = val;
    calc(pos);
  }
  void increment(int pos, T val) {
    pos += n;
    segtree[pos] = join(segtree[pos], val);
    calc(pos);
  }
  T query(int l, int r) {
    T ansl = base;
    T ansr = base;
    l += n;
    r += n + 1;
    while (l < r) {
      if (l & 1) {
        ansl = join(ansl, segtree[l++]);
      }
      if (r & 1) {
        ansr = join(segtree[--r], ansr);
      }
      l >>= 1;
      r >>= 1;
    }
    return join(ansl, ansr);
  }
  T search(T val) {
    assert(__builtin_popcount(n) == 1);
    int idx = 1;
    T curr = base;
    while (idx < n) {
      idx <<= 1;
      if (join(curr, segtree[idx]) < val) {
        curr = join(curr, segtree[idx]);
        ++idx;
      }
    }
    if (join(curr, segtree[idx]) < val) {
      curr = join(curr, segtree[idx]);
      ++idx;
    }
    return idx - n;
  }
};
const int p = 1e9 + 7;
void solve() {
  int n, q;
  cin >> n >> q;
  vector<ll> a(n), w(n);
  cin >> a >> w;
  for (int i = 0; i < n; i++) a[i] -= i;
  vector<ll> aw(n);
  for (int i = 0; i < n; i++) aw[i] = a[i] * w[i], aw[i] %= p;
  int pn = 1;
  while (pn < n) {
    pn <<= 1;
  }
  w.resize(pn);
  aw.resize(pn);
  segment_tree<ll> segtree1(
      w, [&](ll x, ll y) { return x + y; }, 0ll);
  segment_tree<ll> segtree2(
      aw,
      [&](ll x, ll y) {
        ll ans = x + y;
        if (ans >= p) ans -= p;
        return ans;
      },
      0ll);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    if (l > 0) {
      --l;
      --r;
      ll sum = (segtree1.query(l, r) + 1) / 2;
      if (l != 0) sum += segtree1.query(0, l - 1);
      int idx = segtree1.search(sum);
      ll x = a[idx];
      ll ans =
          segtree2.query(idx, r) - segtree2.query(l, idx) +
          (((segtree1.query(l, idx) - segtree1.query(idx, r)) % p) * x) % p;
      ans %= p;
      if (ans < 0) ans += p;
      cout << ans << "\n";
    } else {
      l = -l;
      --l;
      segtree1.set(l, r);
      segtree2.set(l, r * a[l] % p);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}